What Is the Concentration of the First Cation When the Second Starts to Precipitate

Affiliate 15. Equilibria of Other Reaction Classes

fifteen.one Precipitation and Dissolution

Learning Objectives

By the end of this section, yous volition be able to:

• Write chemical equations and equilibrium expressions representing solubility equilibria
• Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations

The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of difficult h2o in your home's water supply are just a few of the many tasks that involve decision-making the equilibrium betwixt a slightly soluble ionic solid and an aqueous solution of its ions.

In some cases, we want to prevent dissolution from occurring. Molar decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca_five(PO_four)₃(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the disuse. On the other hand, sometimes we want a substance to dissolve. Nosotros want the calcium carbonate in a chewable antacid to deliquesce because the [latex]\text{CO}_3^{\;\;2-}[/latex] ions produced in this procedure help soothe an upset stomach.

In this section, nosotros will find out how nosotros tin can control the dissolution of a slightly soluble ionic solid by the application of Le Châtelier's principle. We volition besides learn how to utilize the equilibrium constant of the reaction to make up one's mind the concentration of ions present in a solution.

The Solubility Production Constant

Silver chloride is what's known equally a sparingly soluble ionic solid (Figure 1). Recall from the solubility rules in an earlier chapter that halides of Ag⁺ are not commonly soluble. However, when we add together an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag⁺ and Cl^– ions in equilibrium with undissolved silverish chloride:

[latex]\text{AgCl}(s)\;\underset{\text{precipitation}}{\overset{\text{dissolution}}{\rightleftharpoons}}\;\text{Ag}^{+}(aq)\;+\;\text{Cl}^{-}(aq)[/latex]

This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, simply at the same time, Ag⁺ and Cl^– ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.

Figure 1. Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag⁺ and Cl^– ions in equilibrium with undissolved silver chloride.

The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the solubility production (K _sp) of the solid. Recall from the affiliate on solutions and colloids that we utilise an ion's concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:

[latex]\text{AgCl}(s)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{Cl}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}(aq)][\text{Cl}^{-}(aq)][/latex]

When looking at dissolution reactions such as this, the solid is listed as a reactant, whereas the ions are listed equally products. The solubility product constant, as with every equilibrium constant expression, is written as the product of the concentrations of each of the ions, raised to the ability of their stoichiometric coefficients. Here, the solubility product abiding is equal to Ag⁺ and Cl^– when a solution of silver chloride is in equilibrium with undissolved AgCl. There is no denominator representing the reactants in this equilibrium expression since the reactant is a pure solid; therefore [AgCl] does not appear in the expression for Chiliad _sp.

Some common solubility products are listed in Table 1 according to their 1000 _sp values, whereas a more than extensive compilation of products appears in Appendix J. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small K _sp represents a system in which the equilibrium lies to the left, and so that relatively few hydrated ions would be present in a saturated solution.

Substance	K sp at 25 °C
CuCl	1.2 × 10–half-dozen
CuBr	6.27 × 10–9
AgI	1.5 × ten–16
PbS	7 × ten–29
Al(OH)3	2 × 10–32
Iron(OH)iii	4 × 10–38
Tabular array ane. Common Solubility Products by Decreasing Equilibrium Constants

Example 1

Writing Equations and Solubility Products
Write the ionic equation for the dissolution and the solubility production expression for each of the following slightly soluble ionic compounds:

(a) AgI, silver iodide, a solid with antiseptic properties

(b) CaCO₃, calcium carbonate, the active ingredient in many over-the-counter chewable antacids

(c) Mg(OH)₂, magnesium hydroxide, the active ingredient in Milk of Magnesia

(d) Mg(NH₄)PO₄, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium

(e) Ca₅(PO₄)_iiiOH, the mineral apatite, a source of phosphate for fertilizers

(Hint: When determining how to break (d) and (e) upwardly into ions, refer to the listing of polyatomic ions in the section on chemical classification.)

Solution
(a) [latex]\text{AgI}(s)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{I}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}][\text{I}^{-}][/latex]

(b) [latex]\text{CaCO}_3(due south)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;\text{CO}_3^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{2+}][\text{CO}_3^{\;\;2-}][/latex]

(c) [latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{ii+}(aq)\;+\;ii\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^{-}]^ii[/latex]

(d) [latex]\text{Mg(NH}_4)\text{PO}_4(south)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{PO}_4^{\;\;3-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{two+}][\text{NH}_4^{\;\;+}][\text{PO}_4^{\;\;3-}][/latex]

(eastward) [latex]\text{Ca}_5(\text{PO}_4)iii\text{OH}(s)\;{\rightleftharpoons}\;5\text{Ca}^{2+}(aq)\;+\;iii\text{PO}_4^{\;\;three-}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{2+}]^v[\text{PO}_4^{\;\;3-}]^3[\text{OH}^{-}][/latex]

Check Your Learning
Write the ionic equation for the dissolution and the solubility product for each of the post-obit slightly soluble compounds:

(a) BaSO₄

(b) Ag₂SO_four

(c) Al(OH)_three

(d) Pb(OH)Cl

Answer:

(a) [latex]\text{BaSO}_4(s)\;{\rightleftharpoons}\;\text{Ba}^{2+}(aq)\;+\;\text{SO}_4^{\;\;two-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ba}^{2+}][\text{And then}_4^{\;\;2-}][/latex]; (b) [latex]\text{Ag}_2\text{SO}_4(southward)\;{\rightleftharpoons}\;2\text{Ag}^{+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}]^2[\text{SO}_4^{\;\;2-}][/latex]; (c) [latex]\text{Al(OH)}_3(s)\;{\rightleftharpoons}\;\text{Al}^{2+}(aq)\;+\;3\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Al}^{three+}][\text{OH}^{-}]^3[/latex]; (d) [latex]\text{Atomic number 82(OH)Cl}(s)\;{\rightleftharpoons}\;\text{Atomic number 82}^{2+}(aq)\;+\;\text{OH}^{-}(aq)\;+\;\text{Cl}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Pb}^{2+}][\text{OH}^{-}][\text{Cl}^{-}][/latex]

At present we will extend the discussion of K _sp and show how the solubility product constant is determined from the solubility of its ions, besides as how K _sp can be used to make up one's mind the molar solubility of a substance.

K _sp and Solubility

Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We tin can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the just significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:

[latex]\text{Yard}_p\text{X}_q(s)\;{\rightleftharpoons}\;p\text{M}^{\text{thou+}}(aq)\;+\;q\text{X}^{\text{northward-}}(aq)[/latex]

In this example, we calculate the solubility production by taking the solid's solubility expressed in units of moles per liter (mol/50), known as its molar solubility.

Case ii

Adding of K _sp from Equilibrium Concentrations
We began the chapter with an informal discussion of how the mineral fluorite (Introduction to Chapter 15) is formed. Fluorite, CaF_two, is a slightly soluble solid that dissolves according to the equation:

[latex]\text{CaF}_2(s)\;{\rightleftharpoons}\;\text{Ca}^{ii+}(aq)\;+\;two\text{F}^{-}(aq)[/latex]

The concentration of Ca²⁺ in a saturated solution of CaF_two is 2.xv × ten^–4 Thousand; therefore, that of F^– is 4.xxx × 10^–four M, that is, twice the concentration of Ca²⁺. What is the solubility product of fluorite?

Solution
First, write out the Thousand _sp expression, then substitute in concentrations and solve for G _sp:

[latex]\text{CaF}_2(southward)\;{\rightleftharpoons}\;\text{Ca}^{two+}(aq)\;+\;2\text{F}^{-}(aq)[/latex]

A saturated solution is a solution at equilibrium with the solid. Thus:

[latex]K_{\text{sp}} = [\text{Ca}^{two+}][\text{F}^{-}]^ii = (two.ane\;\times\;10^{-4})(iv.two\;\times\;10^{-4})^ii = 3.7\;\times\;10^{-eleven}[/latex]

As with other equilibrium constants, we do not include units with Thousand _sp.

Check Your Learning
In a saturated solution that is in contact with solid Mg(OH)₂, the concentration of Mgⁱⁱ⁺ is ane.31 × ten^–four M. What is the solubility product for Mg(OH)₂?

[latex]\text{Mg(OH)}_2(due south)\;{\rightleftharpoons}\;\text{Mg}^{two+}(aq)\;+\;2\text{OH}^{-}(aq)[/latex]

Example three

Conclusion of Molar Solubility from K _sp
The K _sp of copper(I) bromide, CuBr, is 6.3 × 10^–9. Calculate the molar solubility of copper bromide.

Solution
The solubility product abiding of copper(I) bromide is six.three × 10^–9.

The reaction is:

[latex]\text{CuBr}(s)\;{\rightleftharpoons}\;\text{Cu}^{+}(aq)\;+\;\text{Br}^{-}(aq)[/latex]

First, write out the solubility product equilibrium constant expression:

[latex]K_{\text{sp}} = [\text{Cu}^{+}][\text{Br}^{-}][/latex]

Create an Water ice table (as introduced in the affiliate on fundamental equilibrium concepts), leaving the CuBr column empty as information technology is a solid and does not contribute to the K _sp:

At equilibrium:

[latex]K_{\text{sp}} = [\text{Cu}^{+}][\text{Br}^{-}][/latex]

[latex]half dozen.iii\;\times\;10^{-9} = (x)(x) = x^2[/latex]

[latex]x = \sqrt{(6.3\;\times\;ten^{-ix})} = seven.9\;\times\;10^{-5}[/latex]

Therefore, the molar solubility of CuBr is 7.ix × 10^–five G.

Bank check Your Learning
The Thou _sp of AgI is 1.5 × 10^–sixteen. Calculate the molar solubility of silver iodide.

Example 4

Conclusion of Tooth Solubility from Grand _sp, Office II
The K _sp of calcium hydroxide, Ca(OH)_ii, is 1.three × 10^{–half-dozen}. Calculate the tooth solubility of calcium hydroxide.

Solution
The solubility product constant of calcium hydroxide is i.iii × 10^{–half-dozen}.

The reaction is:

[latex]\text{Ca(OH)}_2(due south)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;two\text{OH}^{-}(aq)[/latex]

First, write out the solubility product equilibrium constant expression:

[latex]K_{\text{sp}} = [\text{Ca}^{two+}][\text{OH}^{-}]^2[/latex]

Create an ICE table, leaving the Ca(OH)₂ column empty as information technology is a solid and does not contribute to the Yard _sp:

At equilibrium:

[latex]K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^{-}]^2[/latex]

[latex]i.three\;\times\;ten^{-vi} = (x)(2x)^ii = (x)(4x^ii) = 4x^3[/latex]

[latex]x = \sqrt[3]{\frac{1.3\;\times\;10^{-6}}{4}} = vi.9\;\times\;10^{-iii}[/latex]

Therefore, the molar solubility of Ca(OH)_ii is 1.three × 10^–ii M.

Check Your Learning
The Chiliad _sp of PbI₂ is ane.four × 10^–8. Calculate the molar solubility of lead(II) iodide.

Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to apply it in the solubility product constant expression. Instance 5 shows how to perform those unit of measurement conversions before determining the solubility product equilibrium.

Case five

Determination of K _sp from Gram Solubility
Many of the pigments used by artists in oil-based paints (Figure two) are sparingly soluble in water. For example, the solubility of the artist's pigment chrome yellow, PbCrO₄, is 4.6 × x^–six g/L. Make up one's mind the solubility product equilibrium constant for PbCrO₄.

Figure 2. Oil paints comprise pigments that are very slightly soluble in h2o. In addition to chrome yellowish (PbCrO₄), examples include Prussian blue (Fe₇(CN)_xviii), the reddish-orange color vermilion (HgS), and green colour veridian (Cr₂O₃). (credit: Sonny Abesamis)

Solution
We are given the solubility of PbCrO_four in grams per liter. If we convert this solubility into moles per liter, we tin can find the equilibrium concentrations of Pb²⁺ and [latex]\text{CrO}_4^{\;\;2-}[/latex], then K _sp:

1. Use the molar mass of PbCrO₄ [latex](\frac{323.2\;\text{m}}{ane\;\text{mol}})[/latex] to convert the solubility of PbCrO₄ in grams per liter into moles per liter:

   [latex][\text{PbCrO}_4] = \frac{4.vi\;\times\;10^{-6}\;\text{g}\;\text{PbCrO}_4}{1\;\text{L}}\;\times\;\frac{1\;\text{mol}\;\text{PbCrO}_4}{323.2\;\text{g\;PbCrO}_4}[/latex]
   [latex]= \frac{one.4\;\times\;x^{-viii}\;\text{mol\;PbCrO}_4}{one\;\text{50}}[/latex]
   [latex]= 1.4\;\times\;10^{-viii}\;M[/latex]

2. The chemical equation for the dissolution indicates that 1 mol of PbCrO_four gives ane mol of Atomic number 82²⁺(aq) and 1 mol of [latex]\text{CrO}_4^{\;\;two-}(aq)[/latex]:

   [latex]\text{PbCrO}_4(s)\;{\rightleftharpoons}\;\text{Pb}^{2+}(aq)\;+\;\text{CrO}_4^{\;\;two-}(aq)[/latex]

   Thus, both [Pb²⁺] and [latex][\text{CrO}_4^{\;\;ii-}][/latex] are equal to the molar solubility of PbCrO_iv:

   [latex][\text{Lead}^{2+}] = [\text{CrO}_4^{\;\;two-}] = 1.four\;\times\;10^{-8}\;G[/latex]

3. Solve. Thou _sp = [Pb²⁺][latex][\text{CrO}_4^{\;\;2-}][/latex] = (1.4 × 10^–8)(1.4 × 10^–8) = ii.0 × x^–sixteen

Check Your Learning
The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is beingness isolated from ores, is 3.46 grams per liter at twenty °C. What is its solubility production?

Case half dozen

Calculating the Solubility of Hg₂Cl₂
Calomel, Hg_twoCl₂, is a compound composed of the diatomic ion of mercury(I), [latex]\text{Hg}_2^{\;\;2+}[/latex], and chloride ions, Cl^–. Although most mercury compounds are at present known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:

[latex]\text{Hg}_2\text{Cl}_2(s)\;{\rightleftharpoons}\;\text{Hg}_2^{\;\;two+}(aq)\;+\;2\text{Cl}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 1.one\;\times\;10^{-eighteen}[/latex]

Calculate the molar solubility of Hg₂Cl₂.

Solution
The molar solubility of Hg_twoCl₂ is equal to the concentration of [latex]\text{Hg}_2^{\;\;two+}[/latex] ions because for each 1 mol of Hg₂Cl₂ that dissolves, 1 mol of [latex]\text{Hg}_2^{\;\;two+}[/latex] forms:

1. Make up one's mind the direction of change. Before whatsoever Hg₂Cl₂ dissolves, Q is naught, and the reaction volition shift to the right to accomplish equilibrium.
2. Determine x and equilibrium concentrations. Concentrations and changes are given in the following Ice table:
   
   

   Notation that the change in the concentration of Cl^– (twox) is twice every bit large as the change in the concentration of [latex]\text{Hg}_2^{\;\;2+}[/latex] (ten) because 2 mol of Cl^– forms for each 1 mol of [latex]\text{Hg}_2^{\;\;two+}[/latex] that forms. Hg₂Cl₂ is a pure solid, so it does not announced in the calculation.

3. Solve for ten and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for K _sp and calculate the value of x:

   [latex]K_{\text{sp}} = [\text{Hg}_2^{\;\;2+}][\text{Cl}^{-}]^2[/latex]

   [latex]ane.i\;\times\;10^{-18} = (x)(2x)^2[/latex]

   [latex]4x^3 = 1.one\;\times\;ten^{-eighteen}[/latex]

   [latex]x = \sqrt[3]{(\frac{1.1\;\times\;10^{-18}}{iv})} = 6.five\;\times\;10^{-7}\;M[/latex]

   [latex][\text{Hg}_2^{\;\;ii+}] = six.5\;\times\;x^{-7}\;M = 6.5\;\times\;10^{-7}\;M[/latex]

   [latex][\text{Cl}^{-}] = 2x = 2(6.5\;\times\;ten^{-7}) = i.3\;\times\;x^{-six}\;M[/latex]

   The molar solubility of Hg₂Cl_two is equal to [latex][\text{Hg}_2^{\;\;2+}][/latex], or 6.v × 10^–7 M.

4. Check the work. At equilibrium, Q = K _sp:

   [latex]Q = [\text{Hg}_2^{\;\;ii+}][\text{Cl}^{-}]^2 = (6.5\;\times\;x^{-7})(1.3\;\times\;10^{-6})^2 = 1.one\;\times\;10^{-18}[/latex]

   The calculations bank check.

Cheque Your Learning
Determine the tooth solubility of MgF_two from its solubility product: K _sp = 6.4 × x^–9.

Tabulated K _sp values can also be compared to reaction quotients calculated from experimental information to tell whether a solid will precipitate in a reaction under specific conditions: Q equals K _sp at equilibrium; if Q is less than G _sp, the solid will dissolve until Q equals K _sp; if Q is greater than K _sp, atmospheric precipitation will occur at a given temperature until Q equals K _sp.

Using Barium Sulfate for Medical Imaging

Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium chemical compound before taking an 10-ray image. A break of barium sulfate, a chalky powder, is ingested by the patient. Since the K _sp of barium sulfate is one.one × 10^–ten, very little of it dissolves as it coats the lining of the patient's intestinal tract. Barium-coated areas of the digestive tract and then appear on an X-ray as white, allowing for greater visual particular than a traditional X-ray (Figure 3).

Figure 3. The suspension of barium sulfate coats the intestinal tract, which allows for greater visual detail than a traditional X-ray. (credit modification of piece of work by "glitzy queen00"/Wikimedia Commons)

Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous 10-ray is passed through the body so the doctor tin can monitor, on a Television set or estimator screen, the barium sulfate's move as information technology passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux illness, Crohn'due south disease, and ulcers in improver to other conditions.

Visit this website for more information on how barium is used in medical diagnoses and which weather it is used to diagnose.

Predicting Precipitation

The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:

[latex]\text{CaCO}_3(s)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;\text{CO}_3^{\;\;two-}(aq)[/latex]

We can establish this equilibrium either by adding solid calcium carbonate to h2o or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction caliber [latex](\text{Q} = [\text{Ca}^{2+}][\text{CO}_3^{\;\;2-}])[/latex] is equal to the solubility product (K _sp = 8.seven × 10^–9). If we mix a solution of calcium nitrate, which contains Ca²⁺ ions, with a solution of sodium carbonate, which contains [latex]\text{CO}_3^{\;\;2-}[/latex] ions, the slightly soluble ionic solid CaCO₃ will precipitate, provided that the concentrations of Ca^two+ and [latex]\text{CO}_3^{\;\;2-}[/latex] ions are such that Q is greater than G _sp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals M _sp. A saturated solution in equilibrium with the undissolved solid will outcome. If the concentrations are such that Q is less than K _sp, then the solution is not saturated and no precipitate will form.

Nosotros can compare numerical values of Q with G _sp to predict whether atmospheric precipitation volition occur, equally Instance 7 shows. (Notation: Since all forms of equilibrium constants are temperature dependent, we will presume a room temperature environment going forward in this chapter unless a unlike temperature value is explicitly specified.)

Example vii

Precipitation of Mg(OH)₂
The start pace in the preparation of magnesium metal is the precipitation of Mg(OH)₂ from sea water by the improver of lime, Ca(OH)₂, a readily bachelor inexpensive source of OH^– ion:

[latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{ii+}(aq)\;+\;2\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 8.9\;\times\;ten^{-12}[/latex]

The concentration of Mg^two+(aq) in sea water is 0.0537 M. Will Mg(OH)₂ precipitate when enough Ca(OH)₂ is added to requite a [OH^–] of 0.0010 M?

Solution
This problem asks whether the reaction:

[latex]\text{Mg(OH)}_2(s)\;{\leftrightharpoons}\;\text{Mg}^{ii+}(aq)\;+\;2\text{OH}^{-}(aq)[/latex]

shifts to the left and forms solid Mg(OH)_ii when [Mg^two+] = 0.0537 Yard and [OH^–] = 0.0010 M. The reaction shifts to the left if Q is greater than K _sp. Calculation of the reaction quotient nether these conditions is shown here:

[latex]Q = [\text{Mg}^{2+}][\text{OH}^{-}]^2 = (0.0537)(0.0010)^2 = 5.4\;\times\;x^{-8}[/latex]

Because Q is greater than Chiliad _sp (Q = v.4 × 10^–8 is larger than Thousand _sp = 8.9 × 10^–12), we can look the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)_ii(southward) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to One thousand _sp.

Check Your Learning
Use the solubility product in Appendix J to determine whether CaHPO_four volition precipitate from a solution with [Caⁱⁱ⁺] = 0.0001 Grand and [latex][\text{HPO}_4^{\;\;2-}][/latex] = 0.001 Yard.

Respond:

No precipitation of CaHPO₄; Q = 1 × 10^–7, which is less than K _sp

Case 8

Precipitation of AgCl upon Mixing Solutions
Does silvery chloride precipitate when equal volumes of a ii.0 × 10^–4–M solution of AgNO₃ and a ii.0 × ten^–four–M solution of NaCl are mixed?

(Note: The solution also contains Na⁺ and [latex]\text{NO}_3^{\;\;-}[/latex] ions, but when referring to solubility rules, i tin can see that sodium nitrate is very soluble and cannot course a precipitate.)

Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:

[latex]\text{AgCl}(south)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{Cl}^{-}(aq)[/latex]

The solubility product is 1.half-dozen × 10^–x (come across Appendix J).

AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO₃ and NaCl is greater than G _sp. The volume doubles when we mix equal volumes of AgNO_iii and NaCl solutions, then each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag⁺] and [Cl^–] are both equal to:

[latex]\frac{1}{2}(ii.0\;\times\;x^{-4})\;M = 1.0\;\times\;10^{-4}\;K[/latex]

The reaction quotient, Q, is momentarily greater than Yard _sp for AgCl, so a supersaturated solution is formed:

[latex]Q = [\text{Ag}^{+}][\text{Cl}^{-}] = (ane.0\;\times\;10^{-4})(1.0\;\times\;ten^{-4}) = 1.0\;\times\;10^{-8}\;{\textgreater}\;K_{\text{sp}}[/latex]

Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to M _sp.

Check Your Learning
Volition KClO₄ precipitate when 20 mL of a 0.050-M solution of Thousand⁺ is added to eighty mL of a 0.50-Thou solution of [latex]\text{ClO}_4^{\;\;-}[/latex]? (Recollect to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)

Answer:

No, Q = iv.0 × 10^–iii, which is less than K _sp = 1.05 × ten^–two

In the previous two examples, nosotros accept seen that Mg(OH)₂ or AgCl precipitate when Q is greater than One thousand _sp. In general, when a solution of a soluble common salt of the M^m+ ion is mixed with a solution of a soluble salt of the X^n– ion, the solid, M _p X _q precipitates if the value of Q for the mixture of M^{one thousand+} and X^n– is greater than Yard _sp for M _p X _q . Thus, if nosotros know the concentration of i of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we tin can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will presume that precipitation begins when the reaction caliber becomes equal to the solubility product constant.

Example 9

Precipitation of Calcium Oxalate
Blood will non clot if calcium ions are removed from its plasma. Some blood drove tubes contain salts of the oxalate ion, [latex]\text{C}_2\text{O}_4^{\;\;2-}[/latex], for this purpose (Figure 4). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC_twoO₄·H₂O (which also contains water bound in the solid). The concentration of Ca^two+ in a sample of blood serum is 2.ii × 10^–3 M. What concentration of [latex]\text{C}_2\text{O}_4^{\;\;2-}[/latex] ion must be established earlier CaC₂O₄·H₂O begins to precipitate?

Figure 4. Anticoagulants tin can exist added to blood that will combine with the Ca²⁺ ions in claret serum and prevent the blood from clotting. (credit: modification of piece of work by Neeta Lind)

Solution
The equilibrium expression is:

[latex]\text{CaC}_2\text{O}_4(s)\;{\rightleftharpoons}\;\text{Ca}^{two+}(aq)\;+\;\text{C}_2\text{O}_4^{\;\;2-}(aq)[/latex]

For this reaction:

[latex]K_{\text{sp}} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{\;\;2-}] = 1.96\;\times\;10^{-eight}[/latex]

(meet Appendix J)

CaC₂O₄ does non appear in this expression because information technology is a solid. Water does non appear considering it is the solvent.

Solid CaC_twoO₄ does not begin to class until Q equals K _sp. Because we know K _sp and [Ca²⁺], we can solve for the concentration of [latex]\text{C}_2\text{O}_4^{\;\;2-}[/latex] that is necessary to produce the first trace of solid:

[latex]Q = K_{\text{sp}} = [\text{Ca}^{two+}][\text{C}_2\text{O}_4^{\;\;2-}] = 1.96\;\times\;10^{-8}[/latex]

[latex](2.2\;\times\;10^{-3})[\text{C}_2\text{O}_4^{\;\;2-}] = 1.96\;\times\;10^{-8}[/latex]

[latex][\text{C}_2\text{O}_4^{\;\;2-}] = \frac{1.96\;\times\;x^{-8}}{2.ii\;\times\;x^{-3}} = 8.9\;\times\;10^{-6}[/latex]

A concentration of [latex][\text{C}_2\text{O}_4^{\;\;ii-}][/latex] = 8.9 × 10^–6 Yard is necessary to initiate the precipitation of CaC₂O₄ nether these conditions.

Bank check Your Learning
If a solution contains 0.0020 mol of [latex]\text{CrO}_4^{\;\;2-}[/latex] per liter, what concentration of Ag⁺ ion must be reached by adding solid AgNO₃ before Ag₂CrO₄ begins to precipitate? Neglect any increment in volume upon adding the solid silver nitrate.

Information technology is sometimes useful to know the concentration of an ion that remains in solution afterward precipitation. We can use the solubility product for this calculation too: If we know the value of K _sp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same blazon as that in Instance nine—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. Even so, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.

Case 10

Concentrations Following Precipitation
Wearable washed in water that has a manganese [Mn²⁺(aq)] concentration exceeding 0.i mg/L (1.8 × 10^{–half-dozen} K) may be stained by the manganese upon oxidation, but the corporeality of Mn²⁺ in the h2o can be reduced by adding a base of operations. If a person doing laundry wishes to add together a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)_two, what pH is required to go along [Mn²⁺] equal to 1.eight × x^–6 Yard?

Solution
The dissolution of Mn(OH)₂ is described past the equation:

[latex]\text{Mn(OH)}_2(southward)\;{\rightleftharpoons}\;\text{Mn}^{2+}(aq)\;+\;2\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 2\;\times\;10^{-3}[/latex]

We demand to calculate the concentration of OH^– when the concentration of Mnⁱⁱ⁺ is 1.8 × 10^–6 Thou. From that, we calculate the pH. At equilibrium:

[latex]K_{\text{sp}} = [\text{Mn}^{2+}][\text{OH}^{-}]^2[/latex]

or

[latex](i.8\;\times\;10^{-half-dozen})[\text{OH}^{-}]^2 = 2\;\times\;10^{-iii}[/latex]

then

[latex][\text{OH}^{-}] = 3.3\;\times\;ten^{-4}\;M[/latex]

At present we calculate the pH from the pOH:

[latex]\text{pOH} = -\text{log}[\text{OH}^{-}] = -\text{log}(3.iii\;\times\;10\;-\;4) = three.48[/latex]
[latex]\text{pH} = xiv.00\;-\;\text{pOH} = fourteen.00\;-\;three.80 = 10.52[/latex]

If the person doing laundry adds a base, such as the sodium silicate (Na₄SiO₄) in some detergents, to the wash h2o until the pH is raised to 10.52, the manganese ion will exist reduced to a concentration of 1.8 × 10^{–half-dozen} M; at that concentration or less, the ion will non stain clothing.

Bank check Your Learning
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)₂ from sea water by the add-on of Ca(OH)₂. The concentration of Mgⁱⁱ⁺(aq) in sea water is v.37 × x^–2 M. Calculate the pH at which [Mgⁱⁱ⁺] is diminished to 1.0 × x^–v M by the addition of Ca(OH)_two.

Due to their low-cal sensitivity, mixtures of silver halides are used in fiber eyes for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic flick. Even though AgCl (Grand _sp = 1.6 × 10^–10), AgBr (One thousand _sp = 5.0 × 10^–thirteen), and AgI (K _sp = one.5 × 10^–16) are each quite insoluble, we cannot fix a homogeneous solid mixture of them by adding Ag⁺ to a solution of Cl^–, Br^–, and I^–; essentially all of the AgI volition precipitate before whatsoever of the other solid halides form because of its smaller value for K _sp. All the same, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl^–, Br^–, and I^– to a solution of Ag⁺.

When two anions grade slightly soluble compounds with the aforementioned cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller K _sp) generally precipitates first when nosotros add a precipitating agent to a solution containing both anions (or both cations). When the K _sp values of the two compounds differ past two orders of magnitude or more (e.g., ten^–ii vs. 10^–4), almost all of the less soluble compound precipitates before any of the more soluble 1 does. This is an case of selective precipitation, where a reagent is added to a solution of dissolved ions causing 1 of the ions to precipitate out before the rest.

The Part of Precipitation in Wastewater Handling

Solubility equilibria are useful tools in the handling of wastewater carried out in facilities that may treat the municipal water in your city or town (Effigy 5). Specifically, selective atmospheric precipitation is used to remove contaminants from wastewater before it is released dorsum into natural bodies of water. For example, phosphate ions [latex](\text{PO}_4^{\;\;2-})[/latex] are ofttimes present in the water discharged from manufacturing facilities. An abundance of phosphate causes backlog algae to abound, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.

Figure 5. Wastewater handling facilities, such as this one, remove contaminants from wastewater before the h2o is released back into the natural environment. (credit: "eutrophication&hypoxia"/Wikimedia Eatables)

One common fashion to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)₂. The lime is converted into calcium carbonate, a strong base, in the h2o. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca₅(PO4)₃(OH), which then precipitates out of the solution:

[latex]5\text{Ca}^{2+}\;+\;iii\text{PO}_4^{\;\;3-}\;+\;\text{OH}^{-}\;{\leftrightharpoons}\;\text{Ca}_{10}(\text{PO}_4)_6{\cdot}(\text{OH})_2(southward)[/latex]

The precipitate is and so removed by filtration and the water is brought back to a neutral pH by the addition of CO₂ in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(3) chloride and aluminum sulfate.

View this site for more information on how phosphorus is removed from wastewater.

Selective precipitation can as well be used in qualitative analysis. In this method, reagents are added to an unknown chemic mixture in society to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the add-on of the reagent tin can exist used to determine whether the ion is nowadays in the solution.

View this simulation to study the process of salts dissolving and forming saturated solutions and precipitates for specific compounds, or compounds for which you lot select the charges on the ions and the K _sp

Case 11

Precipitation of Silver Halides
A solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO₃ is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

Solution
The two equilibria involved are:

[latex]\text{AgCl}(due south)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{Cl}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 1.6\;\times\;10^{-x}[/latex]

[latex]\text{AgI}(s)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{I}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 1.five\;\times\;x^{-16}[/latex]

If the solution contained about equal concentrations of Cl^– and I^–, and then the silverish common salt with the smallest K _sp (AgI) would precipitate outset. The concentrations are not equal, yet, so we should notice the [Ag⁺] at which AgCl begins to precipitate and the [Ag⁺] at which AgI begins to precipitate. The table salt that forms at the lower [Ag⁺] precipitates first.

For AgI: AgI precipitates when Q equals K _sp for AgI (1.v × 10^–16). When [I^–] = 0.0010 M:

[latex]Q = [\text{Ag}^{+}][\text{I}^{-}] = [\text{Ag}^{+}](0.0010) = 1.v\;\times\;10^{-xvi}[/latex]

[latex][\text{Ag}^{+}] = \frac{ane.8\;\times\;10^{-10}}{0.x} = 1.6\;\times\;10^{-9}[/latex]

AgI begins to precipitate when [Ag⁺] is ane.5 × 10^–13 Thou.

For AgCl: AgCl precipitates when Q equals Grand _sp for AgCl (ane.6 × 10^–10). When [Cl^–] = 0.10 M:

[latex]Q_{\text{sp}} = [\text{Ag}^{+}][\text{Cl}^{-}] = [\text{Ag}^{+}](0.10) = one.6\;\times\;10^{-x}[/latex]

[latex][\text{Ag}^{+}] = \frac{1.eight\;\times\;10^{-x}}{0.10} = 1.6\;\times\;ten^{-9}\;K[/latex]

AgCl begins to precipitate when [Ag⁺] is 1.6 × 10^–ix M.

AgI begins to precipitate at a lower [Ag⁺] than AgCl, so AgI begins to precipitate kickoff.

Check Your Learning
If silverish nitrate solution is added to a solution which is 0.050 M in both Cl^– and Br^– ions, at what [Ag⁺] would atmospheric precipitation begin, and what would exist the formula of the precipitate?

Answer:

[Ag⁺] = 1.0 × 10^–xi M; AgBr precipitates first

Mutual Ion Result

As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acerb acid decreases when the strong electrolyte sodium acetate, NaCH_iiiCO_two, is added. We can explain this effect using Le Châtelier's principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of [latex]\text{H}_3\text{O}^{+}[/latex] to compensate for the increased acetate ion concentration. This increases the concentration of CH₃CO₂H:

[latex]\text{CH}_3\text{CO}_2\text{H}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{CH}_3\text{CO}_2^{\;\;-}[/latex]

Because sodium acetate and acetic acid have the acetate ion in mutual, the influence on the equilibrium is chosen the common ion effect.

The mutual ion outcome can as well have a direct effect on solubility equilibria. Suppose nosotros are looking at the reaction where silver iodide is dissolved:

[latex]\text{AgI}(due south)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{I}^{-}(aq)[/latex]

If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silvery iodide. Le Châtelier'due south principle tells u.s. that when a modify is made to a system at equilibrium, the reaction will shift to counteract that change. In this case, in that location would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.

View this simulation to run across how the common ion effect piece of work with different concentrations of salts.

Example 12

Mutual Ion Event
Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr_two). The 1000 _sp of CdS is 1.0 × 10^–28.

Solution
The get-go thing you lot should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. Nosotros need to use an Ice tabular array to set up this problem and include the CdBr_two concentration as a contributor of cadmium ions:

[latex]\text{CdS}(s)\;{\leftrightharpoons}\;\text{Cd}^{2+}(aq)\;+\;\text{S}^{2-}(aq)[/latex]

[latex]K_{\text{sp}} = [\text{Cd}^{ii+}][\text{S}^{2-}] = ane.0\;\times\;10^{-28}[/latex]

[latex](0.010\;+\;x)(x) = 1.0\;\times\;10^{-28}[/latex]

[latex]x^ii\;+\;0.010x\;-\;1.0\;\times\;10^{-28} = 0[/latex]

We can solve this equation using the quadratic formula, just we tin can also make an assumption to brand this adding much simpler. Since the M _sp value is so small compared with the cadmium concentration, nosotros tin assume that the change betwixt the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our G _sp expression, we would now become:

[latex]K_{\text{sp}} = [\text{Cd}^{2+}][\text{Due south}^{2-}] = 1.0\;\times\;ten^{-28}[/latex]

[latex](0.010)(x) = i.0\;\times\;10^{-28}[/latex]

[latex]10 = 1.0\;\times\;x^{-26}[/latex]

Therefore, the molar solubility of CdS in this solution is ane.0 × 10^–26 M.

Check Your Learning
Summate the molar solubility of aluminum hydroxide, Al(OH)₃, in a 0.015-Yard solution of aluminum nitrate, Al(NO₃)₃. The K _sp of Al(OH)_iii is 2 × 10^–32.

Key Concepts and Summary

The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Thousand _sp, of the solid. When nosotros have a heterogeneous equilibrium involving the slightly soluble solid M _p X _q and its ions M^grand+ and X^n–:

[latex]\text{Thou}_p\text{X}_q(s)\;{\leftrightharpoons}\;p\text{One thousand}^{\text{m}+}(aq)\;+\;q\text{X}^{\text{n}-}(aq)[/latex]

Nosotros write the solubility product expression as:

[latex]K_{\text{sp}} = [\text{G}^{\text{m}+}]^p[\text{X}^{\text{n}-}]^q[/latex]

The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility tin be calculated from its K _sp, provided the just meaning reaction that occurs when the solid dissolves is the germination of its ions.

A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.

A reagent tin exist added to a solution of ions to let one ion to selectively precipitate out of solution. The mutual ion effect tin also play a function in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Châtelier's principle applies and more precipitate comes out of solution then that the molar solubility is reduced.

Primal Equations

• [latex]\text{M}_p\text{X}_q(s)\;{\leftrightharpoons}\;p\text{K}^{\text{thousand}+}(aq)\;+\;q\text{X}^{\text{n}-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{M}^{\text{m}+}]^p[\text{X}^{\text{north}-}]^q[/latex]

Chemical science Cease of Chapter Exercises

1. Complete the changes in concentrations for each of the post-obit reactions:

   (a) [latex]\begin{array}{lccc} \text{AgI}(s)\;{\longrightarrow}\; & \text{Ag}^{+}(aq) & + & \text{I}^{-}(aq) \\[0.5em] & ten & & \rule[0ex]{2.5em}{0.1ex} \cease{array}[/latex]

   (b) [latex]\begin{array}{lccc} \text{CaCO}_3(s)\;{\longrightarrow} & \text{Ca}^{2+}(aq) & + & \text{CO}_3^{\;\;2-}(aq) \\[0.5em] & \dominion[0ex]{2.5em}{0.1ex} & & x \end{assortment}[/latex]

   (c) [latex]\begin{array}{lccc} \text{Mg(OH)}_2(s)\;{\longrightarrow} & \text{Mg}^{2+}(aq) & + & 2\text{OH}^{-}(aq) \\[0.5em] & x & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

   (d) [latex]\brainstorm{array}{lccc} \text{Mg}_3(\text{PO}_4)_2(southward)\;{\longrightarrow} & 3\text{Mg}^{2+}(aq) & + & 2\text{PO}_4^{\;\;iii-}(aq) \\[0.5em] & \rule[0ex]{ii.5em}{0.1ex} & & 2x \cease{assortment}[/latex]

   (e) [latex]\begin{array}{lccccc} \text{Ca}_5(\text{PO}_4)_3\text{OH}(s)\;{\longrightarrow} & 5\text{Ca}^{two+}(aq) & + & 3\text{PO}_4^{\;\;3-}(aq) & + & \text{OH}^{-}(aq) \\[0.5em] & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & x \finish{assortment}[/latex]

2. Complete the changes in concentrations for each of the following reactions:

   (a) [latex]\begin{array}{lccc} \text{BaSO}_4(due south)\;{\longrightarrow} & \text{Ba}^{ii+}(aq) & + & \text{SO}_4^{\;\;2-}(aq) \\[0.5em] & x & & \dominion[0ex]{2.5em}{0.1ex} \end{array}[/latex]

   (b) [latex]\begin{array}{lccc} \text{Ag}_2\text{SO}_4(s)\;{\longrightarrow} & two\text{Ag}^{+}(aq) & + & \text{SO}_4^{\;\;2-}(aq) \\[0.5em] & \dominion[0ex]{2.5em}{0.1ex} & & ten \stop{assortment}[/latex]

   (c) [latex]\brainstorm{assortment}{lccc} \text{Al(OH)}_3(southward)\;{\longrightarrow} & \text{Al}^{3+}(aq) & + & 3\text{OH}^{-}(aq) \\[0.5em] & x & & \rule[0ex]{two.5em}{0.1ex} \end{array}[/latex]

   (d) [latex]\begin{array}{lccccc} \text{Pb(OH)Cl}(s)\;{\longrightarrow} & \text{Pb}^{2+}(aq) & + & \text{OH}^{-}(aq) & + & \text{Cl}^{-}(aq) \\[0.5em] & \dominion[0ex]{ii.5em}{0.1ex} & & x & & \dominion[0ex]{2.5em}{0.1ex} \end{array}[/latex]

   (eastward) [latex]\begin{assortment}{lccc} \text{Ca}_3(\text{AsO}_4)_2(s)\;{\longrightarrow} & 3\text{Ca}^{2+}(aq) & + & 2\text{AsO}_4^{\;\;3-}(aq) \\[0.5em] & 3x & & \rule[0ex]{ii.5em}{0.1ex} \terminate{assortment}[/latex]

3. How do the concentrations of Ag⁺ and [latex]\text{CrO}_4^{\;\;2-}[/latex] in a saturated solution above 1.0 chiliad of solid Ag₂CrO_four change when 100 one thousand of solid Ag₂CrO_four is added to the system? Explain.
4. How do the concentrations of Pbⁱⁱ⁺ and S^2– change when Thou₂S is added to a saturated solution of PbS?
5. What additional information do nosotros need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions afflicted when the temperature is raised?
6. Which of the post-obit slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO₃, CuI, PbCO₃, PbCl_ii, Tl₂S, KClO₄?
7. Which of the post-obit slightly soluble compounds has a solubility greater than that calculated from its solubility production because of hydrolysis of the anion present: AgCl, BaSO_four, CaF₂, Hg_twoI₂, MnCO_three, ZnS, PbS?
8. Write the ionic equation for dissolution and the solubility product (Thousand _sp) expression for each of the following slightly soluble ionic compounds:

   (a) PbCl₂

   (b) Ag₂S

   (c) Sr_three(PO_four)₂

   (d) SrSO₄

9. Write the ionic equation for the dissolution and the K _sp expression for each of the following slightly soluble ionic compounds:

   (a) LaF_iii

   (b) CaCO₃

   (c) Ag₂So_iv

   (d) Lead(OH)₂

10. The Handbook of Chemical science and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

    (a) BaSiF_half-dozen, 0.026 g/100 mL (contains [latex]\text{SiF}_6^{\;\;2-}[/latex] ions)

    (b) Ce(IO₃)_iv, 1.5 × 10^–2 g/100 mL

    (c) Gd₂(SO₄)_three, iii.98 g/100 mL

    (d) (NH₄)_iiPtBr₆, 0.59 1000/100 mL (contains [latex]\text{PtBr}_6^{\;\;2-}[/latex] ions)

11. The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Considering these compounds are simply slightly soluble, assume that the book does not modify on dissolution and calculate the solubility production for each.

    (a) BaSeO_iv, 0.0118 k/100 mL

    (b) Ba(BrO₃)₂·H₂O, 0.xxx g/100 mL

    (c) NH₄MgAsO_iv·6H₂O, 0.038 g/100 mL

    (d) La_ii(MoO₄)₃, 0.00179 yard/100 mL

12. Use solubility products and predict which of the following salts is the about soluble, in terms of moles per liter, in pure water: CaF_two, Hg₂Cl₂, PbI₂, or Sn(OH)₂.
13. Assuming that no equilibria other than dissolution are involved, summate the tooth solubility of each of the post-obit from its solubility product:

    (a) KHC₄H₄O₆

    (b) PbI₂

    (c) Ag₄[Atomic number 26(CN)_{half dozen}], a salt containing the [latex]\text{Fe(CN)}_4^{\;\;-}[/latex] ion

    (d) Hg₂I_ii

14. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility production:

    (a) Ag₂And so₄

    (b) PbBr₂

    (c) AgI

    (d) CaC₂O_four·H₂O

15. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a mutual ion. Bear witness that changes in the initial concentrations of the mutual ions tin be neglected.

    (a) AgCl(south) in 0.025 Thou NaCl

    (b) CaF₂(due south) in 0.00133 M KF

    (c) Ag_iiAnd so₄(s) in 0.500 L of a solution containing 19.50 g of Thou_twoAnd so_four

    (d) Zn(OH)₂(due south) in a solution buffered at a pH of 11.45

16. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Bear witness that changes in the initial concentrations of the common ions can be neglected.

    (a) TlCl(southward) in 1.250 M HCl

    (b) PbI₂(due south) in 0.0355 Grand CaI_two

    (c) Ag_iiCrO_four(s) in 0.225 L of a solution containing 0.856 thou of K₂CrO₄

    (d) Cd(OH)_ii(s) in a solution buffered at a pH of 10.995

17. Assuming that no equilibria other than dissolution are involved, summate the concentration of all solute species in each of the post-obit solutions of salts in contact with a solution containing a common ion. Testify that it is not advisable to neglect the changes in the initial concentrations of the common ions.

    (a) TlCl(due south) in 0.025 Chiliad TlNO_three

    (b) BaF_two(s) in 0.0313 M KF

    (c) MgC₂O_iv in 2.250 L of a solution containing eight.156 g of Mg(NO_three)_ii

    (d) Ca(OH)₂(southward) in an unbuffered solution initially with a pH of 12.700

18. Explain why the changes in concentrations of the common ions in Chapter 15.1 Chemical science End of Chapter Practise 17 tin be neglected.
19. Explicate why the changes in concentrations of the common ions in Affiliate 15.1 Chemistry End of Chapter Exercise 17 cannot be neglected.
20. Calculate the solubility of aluminum hydroxide, Al(OH)₃, in a solution buffered at pH 11.00.
21. Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is nigh soluble in moles per liter and which is most soluble in grams per liter.
22. Most barium compounds are very poisonous; however, barium sulfate is oftentimes administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 3). This use of BaSO_four is possible because of its low solubility. Calculate the molar solubility of BaSO₄ and the mass of barium present in one.00 L of water saturated with BaSO₄.
23. Public Health Service standards for drinking water set a maximum of 250 mg/Fifty (two.60 × 10^–3 M) of [latex]\text{SO}_4^{\;\;2-}[/latex] because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO₄ ("gyp" water) as a outcome or passing through soil containing gypsum, CaSO_four·2H₂O, see these standards? What is [latex]\text{SO}_4^{\;\;2-}[/latex] in such water?
24. Perform the post-obit calculations:

    (a) Summate [Ag⁺] in a saturated aqueous solution of AgBr.

    (b) What will [Ag⁺] be when enough KBr has been added to brand [Br^–] = 0.050 M?

    (c) What will [Br^–] be when enough AgNO₃ has been added to make [Ag⁺] = 0.020 K?

25. The solubility product of CaSO₄·2H₂O is ii.4 × x^–5. What mass of this salt will dissolve in one.0 L of 0.010 One thousand [latex]\text{SO}_4^{\;\;2-}[/latex]?
26. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the post-obit (see Appendix J for solubility products).

    (a) TlCl

    (b) BaF₂

    (c) Ag₂CrO₄

    (d) CaC₂O_four·H₂O

    (e) the mineral anglesite, PbSO_four

27. Bold that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):

    (a) AgI

    (b) Ag_iiThen_iv

    (c) Mn(OH)₂

    (d) Sr(OH)₂·8H₂O

    (e) the mineral brucite, Mg(OH)₂

28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K _sp for each of the slightly soluble solids indicated:

    (a) AgBr: [Ag⁺] = 5.7 × 10^–7 M, [Br^–] = five.seven × 10^–7 Thou

    (b) CaCO₃: [Ca²⁺] = 5.3 × 10^–three K, [latex][\text{CO}_3^{\;\;two-}][/latex] = 9.0 × ten^–7 M

    (c) PbF_ii: [Leadⁱⁱ⁺] = 2.1 × 10^–3 M, [F^–] = iv.two × 10^–3 M

    (d) Ag₂CrO_iv: [Ag⁺] = five.3 × ten^–5 1000, 3.two × 10^–iii Yard

    (e) InF_three: [In³⁺] = 2.3 × ten^–3 M, [F^–] = vii.0 × 10^–iii G

29. The following concentrations are constitute in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, summate One thousand _sp for each of the slightly soluble solids indicated:

    (a) TlCl: [Tl⁺] = 1.21 × ten^–2 Yard, [Cl^–] = one.2 × 10^–2 M

    (b) Ce(IO₃)₄: [Ce⁴⁺] = 1.8 × 10^–4 1000, [latex][\text{IO}_3^{\;\;-}][/latex] = two.6 × 10^–13 One thousand

    (c) Gd₂(Then₄)₃: [Gdⁱⁱⁱ⁺] = 0.132 M, [latex][\text{Then}_4^{\;\;2-}][/latex] = 0.198 Yard

    (d) Ag_twoThen₄: [Ag⁺] = 2.twoscore × 10^–2 One thousand, [latex][\text{SO}_4^{\;\;2-}][/latex] = 2.05 × ten^–ii M

    (e) BaSO₄: [Ba²⁺] = 0.500 M, [latex][\text{SO}_4^{\;\;2-}][/latex] = ii.sixteen × 10^–10 Thou

30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Thou _sp values.)

    (a) KClO₄: [K⁺] = 0.01 M, [latex][\text{ClO}_4^{\;\;-}][/latex] = 0.01 Yard

    (b) One thousand₂PtCl₆: [1000⁺] = 0.01 K, [latex][\text{PtCl}_6^{\;\;2-}][/latex] = 0.01 M

    (c) PbI₂: [Pb²⁺] = 0.003 One thousand, [I^–] = 1.three × ten^–3 M

    (d) Ag_twoSouth: [Ag⁺] = 1 × 10^–x K, [South^2–] = one × 10^–13 M

31. Which of the post-obit compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for K _sp values.)

    (a) CaCO₃: [Caⁱⁱ⁺] = 0.003 M, [latex][\text{CO}_3^{\;\;2-}][/latex] = 0.003 G

    (b) Co(OH)_two: [Co²⁺] = 0.01 M, [OH^–] = 1 × 10^–7 One thousand

    (c) CaHPO₄: [Caⁱⁱ⁺] = 0.01 K, [latex][\text{HPO}_4^{\;\;2-}][/latex] = ii × 10^–six M

    (d) Pb₃(PO₄)₂: [Pb²⁺] = 0.01 M, [latex][\text{PO}_4^{\;\;iii-}][/latex] = 1 × x^–13 M

32. Calculate the concentration of Tl⁺ when TlCl just begins to precipitate from a solution that is 0.0250 M in Cl^–.
33. Calculate the concentration of sulfate ion when BaSO_four merely begins to precipitate from a solution that is 0.0758 M in Ba²⁺.
34. Summate the concentration of Srⁱⁱ⁺ when SrF₂ starts to precipitate from a solution that is 0.0025 Thousand in F^–.
35. Calculate the concentration of [latex]\text{PO}_4^{\;\;three-}[/latex] when Ag_threePO_four starts to precipitate from a solution that is 0.0125 M in Ag⁺.
36. Summate the concentration of F^– required to begin precipitation of CaF₂ in a solution that is 0.010 M in Ca²⁺.
37. Summate the concentration of Ag⁺ required to begin precipitation of Ag₂CO_iii in a solution that is 2.fifty × 10^–6 M in [latex]\text{CO}_3^{\;\;two-}[/latex].
38. What [Ag⁺] is required to reduce [latex][\text{CO}_3^{\;\;2-}][/latex] to 8.2 × 10^–4 K by precipitation of Ag₂CO₃?
39. What [F^–] is required to reduce [Ca²⁺] to 1.0 × ten^–4 Grand by precipitation of CaF₂?
40. A volume of 0.800 L of a two × 10^–4–M Ba(NO₃)₂ solution is added to 0.200 L of 5 × 10^–four M Li_twoAnd then_iv. Does BaSO₄ precipitate? Explain your answer.
41. Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing NiCO₃ be washed to deliquesce 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO₃ (K _sp = ane.36 × 10^–seven).

    (b) If the NiCO_iii were a contaminant in a sample of CoCO₃ (K _sp = 1.0 × 10^–12), what mass of CoCO_iii would take been lost? Proceed in mind that both NiCO₃ and CoCO₃ dissolve in the same solution.

42. Iron concentrations greater than 5.iv × 10^{–half-dozen} M in water used for laundry purposes can cause staining. What [OH^–] is required to reduce [Iron²⁺] to this level by precipitation of Fe(OH)_two?
43. A solution is 0.010 M in both Cu²⁺ and Cd²⁺. What percentage of Cdⁱⁱ⁺ remains in the solution when 99.9% of the Cu²⁺ has been precipitated every bit CuS past calculation sulfide?
44. A solution is 0.15 M in both Atomic number 82²⁺ and Ag⁺. If Cl^– is added to this solution, what is [Ag⁺] when PbCl_two begins to precipitate?
45. What reagent might exist used to divide the ions in each of the post-obit mixtures, which are 0.ane Yard with respect to each ion? In some cases information technology may be necessary to command the pH. (Hint: Consider the K _sp values given in Appendix J.)

    (a) [latex]\text{Hg}_2^{\;\;ii+}[/latex] and Cu²⁺

    (b) [latex]\text{SO}_4^{\;\;two-}[/latex] and Cl^–

    (c) Hg²⁺ and Co²⁺

    (d) Zn^two+ and Sr²⁺

    (e) Ba²⁺ and Mg²⁺

    (f) [latex]\text{CO}_3^{\;\;2-}[/latex] and OH^–

46. A solution contains 1.0 × 10^–5 mol of KBr and 0.ten mol of KCl per liter. AgNO₃ is gradually added to this solution. Which forms outset, solid AgBr or solid AgCl?
47. A solution contains i.0 × 10^–2 mol of KI and 0.10 mol of KCl per liter. AgNO₃ is gradually added to this solution. Which forms starting time, solid AgI or solid AgCl?
48. The calcium ions in man claret serum are necessary for coagulation (Figure 4). Potassium oxalate, K₂C₂O₄, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium equally a precipitate of CaC₂O_four·H₂O. Information technology is necessary to remove all but i.0% of the Ca²⁺ in serum in order to preclude coagulation. If normal claret serum with a buffered pH of 7.40 contains 9.5 mg of Ca^two+ per 100 mL of serum, what mass of K₂C_twoO₄ is required to prevent the coagulation of a 10 mL blood sample that is 55% serum past book? (All volumes are accurate to two significant figures. Notation that the book of serum in a 10-mL blood sample is v.v mL. Assume that the K_sp value for CaC₂O₄ in serum is the same as in water.)
49. About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca₃(PO₄)₂. The normal mid range calcium content excreted in the urine is 0.10 g of Caⁱⁱ⁺ per day. The normal mid range amount of urine passed may be taken as 1.4 L per solar day. What is the maximum concentration of phosphate ion that urine can incorporate before a calculus begins to course?
50. The pH of normal urine is 6.30, and the total phosphate concentration [latex]([\text{PO}_4^{\;\;iii-}]\;+\;[\text{HPO}_4^{\;\;two-}]\;+\;[\text{H}_2\text{PO}_4^{\;\;-}]\;+\;[\text{H}_3\text{PO}_4])[/latex] is 0.020 M. What is the minimum concentration of Ca²⁺ necessary to induce kidney stone formation? (Come across Chemistry Terminate of Chapter Do 49 for additional information.)
51. Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the product of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:

    [latex]\text{Mg}^{ii+}(aq)\;+\;\text{Ca(OH)}_2(aq)\;{\longrightarrow}\;\text{Mg(OH)}_2(s)\;+\;\text{Ca}^{2+}(aq)[/latex]

    [latex]\text{Mg(OH)}_2(south)\;+\;ii\text{HCl}(aq)\;{\longrightarrow}\;\text{MgCl}_2(due south)\;+\;2\text{H}_2\text{O}(l)[/latex]

    [latex]\text{MgCl}_2(l)\;{\xrightarrow{\text{electrolysis}}}\;\text{Mg}(s)\;+\;\text{Cl}_2(g)[/latex]

    Sea water has a density of 1.026 g/cm³ and contains 1272 parts per million of magnesium equally Mgⁱⁱ⁺(aq) by mass. What mass, in kilograms, of Ca(OH)₂ is required to precipitate 99.9% of the magnesium in 1.00 × 10³ Fifty of sea water?

52. Hydrogen sulfide is bubbled into a solution that is 0.10 Yard in both Pb^two+ and Iron^two+ and 0.30 Thou in HCl. Afterward the solution has come to equilibrium information technology is saturated with H_iiS ([H₂S] = 0.10 M). What concentrations of Atomic number 82²⁺ and Fe^two+ remain in the solution? For a saturated solution of H₂South we tin can use the equilibrium:

    [latex]\text{H}_2\text{S}(aq)\;+\;two\text{H}_2\text{O}(fifty)\;{\leftrightharpoons}\;ii\text{H}_3\text{O}^{+}(aq)\;+\;\text{S}^{2-}(aq)\;\;\;\;\;\;\;K = i.0\;\times\;10^{-26}[/latex]

    (Hint: The [latex][\text{H}_3\text{O}^{+}][/latex] changes as metallic sulfides precipitate.)

53. Perform the post-obit calculations involving concentrations of iodate ions:

    (a) The iodate ion concentration of a saturated solution of La(IO_iii)₃ was plant to be 3.1 × 10^–3 mol/L. Find the K _sp.

    (b) Find the concentration of iodate ions in a saturated solution of Cu(IO₃)₂ (K _sp = 7.4 × 10^–8).

54. Calculate the tooth solubility of AgBr in 0.035 M NaBr (K _sp = 5 × 10^–13).
55. How many grams of Pb(OH)₂ will dissolve in 500 mL of a 0.050-M PbCl₂ solution (K _sp = 1.ii × 10^–15)?
56. Employ the simulation from the before Link to Learning to complete the following exercise:. Using 0.01 g CaF_two, give the Thousand_sp values plant in a 0.2-G solution of each of the salts. Talk over why the values modify as you modify soluble salts.
57. How many grams of Milk of Magnesia, Mg(OH)₂ (due south) (58.3 one thousand/mol), would be soluble in 200 mL of h2o. K _sp = 7.1 × 10^–12. Include the ionic reaction and the expression for K _sp in your answer. (K _w = 1 × 10^–14 = [H₃O⁺][OH^–])
58. Two hypothetical salts, LM₂ and LQ, have the same molar solubility in H₂O. If 1000 _sp for LM₂ is 3.20 × 10^–5, what is the One thousand _sp value for LQ?
59. Which of the following carbonates will form commencement? Which of the following volition form last? Explain.

    (a) [latex]\text{MgCO}_3\;\;\;\;\;\;\;K_{\text{sp}} = three.five\;\times\;10^{-viii}[/latex]

    (b) [latex]\text{CaCO}_3\;\;\;\;\;\;\;K_{\text{sp}} = 4.2\;\times\;10^{-7}[/latex]

    (c) [latex]\text{SrCO}_3\;\;\;\;\;\;\;K_{\text{sp}} = 3.9\;\times\;ten^{-9}[/latex]

    (d) [latex]\text{BaCO}_3\;\;\;\;\;\;\;K_{\text{sp}} = 4.4\;\times\;10^{-v}[/latex]

    (e) [latex]\text{MnCO}_3\;\;\;\;\;\;\;K_{\text{sp}} = five.one\;\times\;x^{-ix}[/latex]

60. How many grams of Zn(CN)₂(southward) (117.44 one thousand/mol) would be soluble in 100 mL of H₂O? Include the balanced reaction and the expression for Thou _sp in your answer. The K _sp value for Zn(CN)₂(s) is 3.0 × 10^–16.

Glossary

mutual ion effect

effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a subtract in the ionization of a weak acid or base

molar solubility

solubility of a compound expressed in units of moles per liter (mol/Fifty)

selective atmospheric precipitation

procedure in which ions are separated using differences in their solubility with a given precipitating reagent

solubility production (K _sp)

equilibrium constant for the dissolution of a slightly soluble electrolyte

Solutions

Answers to Chemistry End of Chapter Exercises

1.

(a) [latex]\brainstorm{array}{lccc} \text{AgI}(s)\;{\rightleftharpoons}\; & \text{Ag}^{+}(aq) & + & \text{I}^{-}(aq) \\[0.5em] & ten & & \rule[-0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}x \stop{assortment}[/latex]

(b) [latex]\brainstorm{array}{lccc} \text{CaCO}_3(s)\;{\rightleftharpoons} & \text{Ca}^{2+}(aq) & + & \text{CO}_3^{\;\;ii-}(aq) \\[0.5em] & \rule[-0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}x & & 10 \end{array}[/latex]

(c) [latex]\brainstorm{array}{lccc} \text{Mg(OH)}_2(s)\;{\rightleftharpoons} & \text{Mg}^{2+}(aq) & + & 2\text{OH}^{-}(aq) \\[0.5em] & x & & \dominion[-0.25ex]{1em}{0.1ex}\hspace{-1em}2x \cease{array}[/latex]

(d) [latex]\begin{array}{lccc} \text{Mg}_3(\text{PO}_4)_2(s)\;{\rightleftharpoons} & three\text{Mg}^{2+}(aq) & + & 2\text{PO}_4^{\;\;3-}(aq) \\[0.5em] & \rule[-0.25ex]{1em}{0.1ex}\hspace{-1em}3x & & 2x \end{array}[/latex]

(e) [latex]\brainstorm{array}{lccccc} \text{Ca}_5(\text{PO}_4)_3\text{OH}(s)\;{\rightleftharpoons} & 5\text{Ca}^{2+}(aq) & + & 3\text{PO}_4^{\;\;3-}(aq) & + & \text{OH}^{-}(aq) \\[0.5em] & \dominion[-0.25ex]{1em}{0.1ex}\hspace{-1em}5x & & \rule[-0.25ex]{1em}{0.1ex}\hspace{-1em}3x & & x \end{array}[/latex]

iii. There is no change. A solid has an activeness of 1 whether there is a little or a lot.

5. The solubility of silverish bromide at the new temperature must be known. Commonly the solubility increases and some of the solid silverish bromide will deliquesce.

7. CaF_two, MnCO₃, and ZnS

9. (a) [latex]\text{LaF}_3(s)\;{\rightleftharpoons}\;\text{La}^{3+}(aq)\;+\;3\text{F}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{La}^{iii+}][\text{F}^{-}]^3[/latex];
(b) [latex]\text{CaCO}_3(s)\;{\rightleftharpoons}\;\text{Ca}^{ii+}(aq)\;+\;\text{CO}_3^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{two+}][\text{CO}_3^{\;\;2-}][/latex];
(c) [latex]\text{Ag}_2\text{So}_4(s)\;{\rightleftharpoons}\;2\text{Ag}^{+}(aq)\;+\;\text{Then}_4^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}]^2[\text{And so}_4^{\;\;2-}][/latex];
(d) [latex]\text{Pb(OH)}_2(s)\;{\rightleftharpoons}\;\text{Pb}^{ii+}(aq)\;+\;2\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Pb}^{2+}][\text{OH}^{-}]^2[/latex]

eleven. (a)1.77 × x^–7; (b) 1.half dozen × 10^{–half dozen}; (c) 2.ii × 10^–9; (d) seven.91 × 10^–22

thirteen. (a) 2 × 10^–2 M; (b) ane.5 × 10^–3 M; (c) 2.27 × 10^–ix Grand; (d) ii.2 × x^–10 Thousand

15. (a) half-dozen.4 × ten⁻⁹ M = [Ag⁺], [Cl⁻] = 0.025 M
Bank check: [latex]\frac{six.4\;\times\;10^{-9}\;M}{0.025\;M}\;\times\;100\% = 2.vi\;\times\;ten^{-v}\;%[/latex],an insignificant change;

(b) 2.two × 10⁻⁵ Yard = [Ca²⁺], [F⁻] = 0.0013 Grand
Bank check: [latex]\frac{2.26\;\times\;ten^{-v}\;M}{0.00133\;M}\;\times\;100\% = 1.70\%[/latex]. This value is less than five% and can be ignored.

(c) 0.2238 M = [latex][\text{Then}_4^{\;\;2-}][/latex]; [Ag⁺] = 7.4 × x^–3 Yard
Cheque: [latex]\frac{3.7\;\times\;10^{-3}}{0.2238}\;\times\;100\% = i.64\;\times\;ten^{-2}\%[/latex]; the status is satisfied.

(d) [OH^–] = 2.8 × 10^–three One thousand; 5.seven × 10⁻¹² Yard = [Znⁱⁱ⁺]
Check: [latex]\frac{5.7\;\times\;ten^{-12}}{2.8\;\times\;10^{-3}}\;\times\;100\% = 2.0\;\times\;10^{-7}\%[/latex]; x is less than five% of [OH^–] and is, therefore, negligible.

17. (a) [Cl^–] = 7.vi × 10⁻³ Yard
Check: [latex]\frac{7.vi\;\times\;ten^{-iii}}{0.025}\;\times\;100\% = 30\%[/latex]
This value is too large to drop x. Therefore solve by using the quadratic equation:
[Ti⁺] = 3.1 × ten^–2 M
[Cl^–] = 6.1 × ten^–3

(b) [Ba^two+] = 7.7 × ten^–4 One thousand
Check: [latex]\frac{7.seven\;\times\;ten^{-four}}{0.0313}\;\times\;100\% = ii.4\%[/latex]
Therefore, the condition is satisfied.
[Ba²⁺] = 7.vii × 10^–iv Grand
[F^–] = 0.0321 Yard;

(c) Mg(NO₃)₂ = 0.02444 M
[latex][\text{C}_2\text{O}_4^{\;\;2-}] = 2.ix\;\times\;10^{-5}[/latex]
Check: [latex]\frac{2.9\;\times\;10^{-5}}{0.02444}\;\times\;100\% = 0.12\%[/latex]
The condition is satisfied; the above value is less than v%.
[latex][\text{C}_2\text{O}_4^{\;\;ii-}] = 2.9\;\times\;ten^{-5}\;M[/latex]
[Mg^two+] = 0.0244 Chiliad

(d) [OH^–] = 0.0501 Thousand
[Ca²⁺] = 3.15 × 10^–3
Cheque: [latex]\frac{3.xv\;\times\;ten^{-3}}{0.050}\;\times\;100\% = half dozen.28\%[/latex]
This value is greater than v%, so a more exact method, such as successive approximations, must be used.
[Ca²⁺] = 2.8 × 10^–iii M
[OH^–] = 0.053 × 10^–2 M

xix. The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the alter.

21. CaSO₄∙2H₂O is the most soluble Ca salt in mol/50, and information technology is also the most soluble Ca common salt in g/L.

23. iv.8 × 10^–3 M = [latex][\text{And then}_4^{\;\;ii-}][/latex] = [Ca²⁺]; Since this concentration is higher than 2.60 × x^–3 Grand, "gyp" water does not come across the standards.

25. Mass (CaSO₄·2H₂O) = 0.72 g/Fifty

27. (a) [Ag⁺] = [I^–] = 1.iii × 10^–v Grand; (b) [Ag⁺] = 2.88 × 10^–2 Yard, [latex][\text{So}_4^{\;\;two-}][/latex] = 1.44 × 10^–2 M; (c) [Mn²⁺] = 3.7 × 10^–5 M, [OH^–] = vii.4 × x^–v M; (d) [Sr²⁺] = 4.iii × 10^–ii M, [OH^–] = 8.6 × ten^–2 M; (e) [Mg²⁺] = 1.3 × 10^–4 M, [OH^–] = two.6 × ten^–iv M.

29. (a) 2.0 × x^–four; (b) 5.1 × 10^–17; (c) 1.35 × 10^–4; (d) one.18 × ten^–v; (e) i.08 × 10^–10

31. (a) CaCO₃ does precipitate.
(b) The compound does not precipitate.
(c) The chemical compound does not precipitate.
(d) The chemical compound precipitates.

33. 3.03 × 10^−seven G

35. 9.2 × x^−xiii M

37. [Ag⁺] = 1.8 × 10^–3 M

39. half-dozen.3 × ten^–4

41. (a) two.25 L; (b) vii.2 × 10^–vii g

43. 100% of information technology is dissolved

45. (a) [latex]\text{Hg}_2^{\;\;2+}[/latex] and Cuⁱⁱ⁺: Add [latex]\text{SO}_4^{\;\;2-}[/latex].
(b) [latex]\text{And then}_4^{\;\;2-}[/latex] and Cl^–: Add Baⁱⁱ⁺.
(c) Hg²⁺ and Co²⁺: Add S^2–.
(d) Zn^two+ and Sr^two+: Add OH^– until [OH^–] = 0.050 M.
(east) Ba²⁺ and Mg²⁺: Add [latex]\text{SO}_4^{\;\;2-}[/latex].
(f) [latex]\text{CO}_3^{\;\;2-}[/latex] and OH^–: Add Ba²⁺.

47. AgI will precipitate get-go.

49. one.5 × x⁻¹² M

51. 3.99 kg

53. (a) 3.1 × 10^–xi; (b) [Cu²⁺] = 2.6 × 10^–3; [latex][\text{IO}_3^{\;\;-}][/latex] = five.iii × x^–3

55. 1.8 × 10^–five 1000 Pb(OH)₂

57. [latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}\;+\;2\text{OH}^{-}\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^{-}]^2[/latex]

1.23 × ten⁻³ chiliad Mg(OH)₂

59. MnCO_three will form first, since information technology has the smallest M _sp value it is the least soluble. MnCO₃ will be the final to precipitate, it has the largest M _sp value.

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Source: https://opentextbc.ca/chemistry/chapter/15-1-precipitation-and-dissolution/

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