15.1 Precipitation and Dissolution

The Solubility Product

Silver chloride is what’s known as a sparingly soluble ionic solid (Figure 15.2). Recall from the solubility rules in an earlier chapter that halides of Ag⁺ are not normally soluble. However, when we add 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 silver chloride:

This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but 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 15.2 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 product (K_sp) of the solid. Recall from the chapter on solutions and colloids that we use an ion’s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:

Note that the K_sp expression does not contain a term in the denominator for the concentration of the reactant, AgCl. According to the guidelines for deriving mass-action expressions described in an earlier chapter on equilibrium, only gases and solutes are represented. Solids and liquids are assigned concentration values of one and thus do not appear in equilibrium constant expressions; therefore, [AgCl] does not appear in the expression for K_sp.

Some common solubility products are listed in Table 15.1 according to their K_sp values, whereas a more extensive compilation of solubility 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, so that relatively few hydrated ions would be present in a saturated solution.

Now we will extend the discussion of K_sp and show how the solubility product is determined from the solubility of its ions, as well as how K_sp can be used to determine the molar solubility of a substance.

K_sp and Solubility

The K_sp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:

For cases such as these, one may derive K_sp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.

Example 15.3

Determination of Molar Solubility from K_sp

The K_sp of copper(I) bromide, CuBr, is 6.3 $\times$ 10^–9. Calculate the molar solubility of copper bromide.

Solution

The solubility product of copper(I) bromide is 6.3 $\times$ 10^–9.

The reaction is:

$$\text{CuBr}(s)\rightleftharpoons {\text{Cu}}^{\text{+}}(aq)+{\text{Br}}^{\text{−}}(aq)$$

15.8

First, write out the solubility product expression:

$$K_{\text{sp}}=[{\text{Cu}}^{\text{+}}][{\text{Br}}^{\text{−}}]$$

15.9

Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the K_sp:

At equilibrium:

$$K_{\text{sp}}={[\text{Cu}}^{\text{+}}][{\text{Br}}^{\text{−}}]$$

15.10

$$6.3\times {10}^{-9}=(x)(x)=x^2$$

15.11

$$x=\sqrt{(6.3\times {10}^{-9})}=\text{7.9}\times {10}^{-5}$$

15.12

Therefore, the molar solubility of CuBr is 7.9 $\times$ 10^–5 M.

Check Your Learning

The K_sp of AgI is 1.5 $\times$ 10^–16. Calculate the molar solubility of silver iodide.

Answer:

1.2 $\times$ 10^–8 M

Example 15.4

Determination of Molar Solubility from K_sp, Part II

The K_sp of calcium hydroxide, Ca(OH)₂, is 1.3 $\times$ 10^–6. Calculate the molar solubility of calcium hydroxide.

Solution

The solubility product of calcium hydroxide is 1.3 $\times$ 10^–6.

The reaction is:

$${\text{Ca(OH)}}_2(s)\rightleftharpoons {\text{Ca}}^{\text{2+}}(aq)+{\text{2OH}}^{\text{−}}(aq)$$

15.13

First, write out the solubility product expression:

$$K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^2$$

15.14

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

At equilibrium:

$$K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^2$$

15.15

$$1.3\times {10}^{-6\text{}}=(x)(2x{)}^2=(x)(4x^2)=4x^3$$

15.16

$$x=\sqrt[3]{\frac{1.3\times {10}^{-6}}{4}}=\text{6.9}\times {10}^{-3}$$

15.17

Therefore, the molar solubility of Ca(OH)₂ is 6.9 $\times$ 10^–3 M.

Check Your Learning

The K_sp of PbI₂ is 1.4 $\times$ 10^–8. Calculate the molar solubility of lead(II) iodide.

Answer:

1.5 $\times$ 10^–3 M

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 use it in the solubility product expression. Example 15.5 shows how to perform those unit conversions before determining the solubility product equilibrium.

Example 15.5

Determination of K_sp from Gram Solubility

Many of the pigments used by artists in oil-based paints (Figure 15.3) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO₄, is 4.6 $\times$ 10^–6 g/L. Determine the solubility product for PbCrO₄.

Figure 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO₄), examples include Prussian blue (Fe₇(CN)₁₈), the reddish-orange color vermilion (HgS), and green color veridian (Cr₂O₃). (credit: Sonny Abesamis)

Solution

We are given the solubility of PbCrO₄ in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb²⁺ and ${\text{CrO}}_4{}^{\text{2−}},$ then K_sp:

1. Step 1.

   Use the molar mass of PbCrO₄ $\left(\frac{323.2\text{g}}{1\text{mol}}\right)$ to convert the solubility of PbCrO₄ in grams per liter into moles per liter:

   $$\begin{array}{}\\ \\ \left[{\text{PbCrO}}_4\right]=\frac{4.6\times {10}^{-6}{\text{g PbCrO}}_4}{1\text{L}}\times \frac{1{\text{mol PbCrO}}_4}{323.2{\text{g PbCrO}}_4}\\ =\frac{1.4\times {10}^{-8}{\text{mol PbCrO}}_4}{1\text{L}}\\ =1.4\times {10}^{-8}M\end{array}$$

   15.18
2. Step 2.

   The chemical equation for the dissolution indicates that 1 mol of PbCrO₄ gives 1 mol of Pb²⁺(aq) and 1 mol of ${\text{CrO}}_4{}^{\text{2−}}(aq)\text{:}$

   $${\text{PbCrO}}_4(s)\rightleftharpoons {\text{Pb}}^{\text{2+}}(aq)+{\text{CrO}}_4{}^{\text{2−}}(aq)$$

   15.19

   Thus, both [Pb²⁺] and $[{\text{CrO}}_4{}^{\text{2−}}]$ are equal to the molar solubility of PbCrO₄:

   $$[{\text{Pb}}^{\text{2+}}]=[{\text{CrO}}_4{}^{\text{2−}}]=\text{1.4}\times {10}^{-8}M$$

   15.20
3. Step 3.

   Solve. K_sp = [Pb²⁺]$[{\text{CrO}}_4{}^{\text{2−}}]$ = (1.4 $\times$ 10^–8)(1.4 $\times$ 10^–8) = 2.0 $\times$ 10^–16

Check Your Learning

The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.46 grams per liter at 20 °C. What is its solubility product?

Answer:

2.08 $\times$ 10^–4

Example 15.6

Calculating the Solubility of Hg₂Cl₂

Calomel, Hg₂Cl₂, is a compound composed of the diatomic ion of mercury(I), ${\text{Hg}}_2{}^{\text{2+}},$ and chloride ions, Cl^–. Although most mercury compounds are now 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:

$${\text{Hg}}_2{\text{Cl}}_2(s)\rightleftharpoons {\text{Hg}}_2{}^{\text{2+}}(aq)+{\text{2Cl}}^{\text{−}}(aq)K_{\text{sp}}=\text{1.1}\times {10}^{-18}$$

15.21

Calculate the molar solubility of Hg₂Cl₂.

Solution

The molar solubility of Hg₂Cl₂ is equal to the concentration of ${\text{Hg}}_2{}^{\text{2+}}$ ions because for each 1 mol of Hg₂Cl₂ that dissolves, 1 mol of ${\text{Hg}}_2{}^{\text{2+}}$ forms:

1. Step 1.

   Determine the direction of change. Before any Hg₂Cl₂ dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium.

2. Step 2.

   Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:

   Note that the change in the concentration of Cl^– (2x) is twice as large as the change in the concentration of ${\text{Hg}}_2{}^{\text{2+}}$ (x) because 2 mol of Cl^– forms for each 1 mol of ${\text{Hg}}_2{}^{\text{2+}}$ that forms. Hg₂Cl₂ is a pure solid, so it does not appear in the calculation.

3. Step 3.

   Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for K_sp and calculate the value of x:

   $$K_{\text{sp}}=[{\text{Hg}}_{\text{2}}{}^{\text{2+}}]{[{\text{Cl}}^{\text{−}}]}^2$$

   15.22

   $$1.1\times {10}^{-18}=(x)(2x{)}^2$$

   15.23

   $$4x^3=\text{1.1}\times {10}^{-18}$$

   15.24

   $$x=\sqrt[3]{\left(\frac{1.1\times {10}^{-18}}{4}\right)}=\text{6.5}\times {10}^{-7}M$$

   15.25

   $$[{\text{Hg}}_2{}^{\text{2+}}]=\text{6.5}\times {10}^{-7}M=\text{6.5}\times {10}^{-7}M$$

   15.26

   $$[{\text{Cl}}^{\text{−}}]=2x=\text{2(6.5}\times {10}^{-7})=1.3\times {10}^{-6}M$$

   15.27

   The molar solubility of Hg₂Cl₂ is equal to $[{\text{Hg}}_{\text{2}}{}^{\text{2+}}],$ or 6.5 $\times$ 10^–7 M.

4. Step 4.

   Check the work. At equilibrium, Q = K_sp:

   $$Q=[{\text{Hg}}_{\text{2}}{}^{\text{2+}}]{[{\text{Cl}}^{\text{−}}]}^2=\text{(6.5}\times {10}^{-7}){(1.3\times {10}^{-6})}^2=1.1\times {10}^{-18}$$

   15.28

   The calculations check.

Check Your Learning

Determine the molar solubility of MgF₂ from its solubility product: K_sp = 6.4 $\times$ 10^–9.

Answer:

1.2 $\times$ 10^–3 M

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

How Sciences Interconnect

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 compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the K_sp of barium sulfate is 1.1 $\times$ 10^–10, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 15.4).

Figure 15.4 The suspension of barium sulfate coats the intestinal tract, which allows for greater visual detail than a traditional X-ray. (credit modification of work by “glitzy queen00”/Wikimedia Commons)

Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate’s movement as it passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.

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

Predicting Precipitation

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

We can establish this equilibrium either by adding solid calcium carbonate to water 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 quotient $(\text{Q}=[{\text{Ca}}^{2+}][{\text{CO}}_3{}^{\text{2−}}])$ is equal to the solubility product (K_sp = 8.7 $\times$ 10^–9). If we mix a solution of calcium nitrate, which contains Ca²⁺ ions, with a solution of sodium carbonate, which contains ${\text{CO}}_3{}^{\text{2−}}$ ions, the slightly soluble ionic solid CaCO₃ will precipitate, provided that the concentrations of Ca²⁺ and ${\text{CO}}_3{}^{\text{2−}}$ ions are such that Q is greater than K_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 K_sp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than K_sp, then the solution is not saturated and no precipitate will form.

We can compare numerical values of Q with K_sp to predict whether precipitation will occur, as Example 15.7 shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)

In the previous two examples, we have seen that Mg(OH)₂ or AgCl precipitate when Q is greater than K_sp. In general, when a solution of a soluble salt of the M^m+ ion is mixed with a solution of a soluble salt of the X^n– ion, the solid, M_pX_q precipitates if the value of Q for the mixture of M^m+ and X^n– is greater than K_sp for M_pX_q. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product.

Example 15.9

Precipitation of Calcium Oxalate

Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, ${\text{C}}_2{\text{O}}_4{}^{\text{2−}},$ for this purpose (Figure 15.5). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC₂O₄·H₂O (which also contains water bound in the solid). The concentration of Ca²⁺ in a sample of blood serum is 2.2 $\times$ 10^–3 M. What concentration of ${\text{C}}_2{\text{O}}_4{}^{\text{2−}}$ ion must be established before CaC₂O₄·H₂O begins to precipitate?

Figure 15.5 Anticoagulants can be added to blood that will combine with the Ca²⁺ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)

Solution

The equilibrium expression is:

$${\text{CaC}}_2{\text{O}}_4(s)\rightleftharpoons {\text{Ca}}^{\text{2+}}(aq)+{\text{C}}_2{\text{O}}_4{}^{\text{2−}}(aq)$$

15.36

For this reaction:

$$K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}][{\text{C}}_2{\text{O}}_4{}^{\text{2−}}]=1.96\times {10}^{-8}$$

15.37

(see Appendix J)

CaC₂O₄ does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CaC₂O₄ does not begin to form until Q equals K_sp. Because we know K_sp and [Ca²⁺], we can solve for the concentration of ${\text{C}}_2{\text{O}}_4{}^{\text{2−}}$ that is necessary to produce the first trace of solid:

$$Q=K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}]{[\text{C}}_2{\text{O}}_4{}^{\text{2−}}]=\text{1.96}\times {10}^{-8}$$

15.38

$$(2.2\times {10}^{-3})[{\text{C}}_{\text{2}}{\text{O}}_4{}^{\text{2−}}]=\text{1.96}\times {10}^{-8}$$

15.39

$$[{\text{C}}_2{\text{O}}_4{}^{\text{2−}}]=\frac{1.96\times {10}^{-8}}{2.2\times {10}^{-3}}=\text{8.9}\times {10}^{-6}$$

15.40

A concentration of $[{\text{C}}_2{\text{O}}_4{}^{\text{2−}}]$ = 8.9 $\times$ 10^–6 M is necessary to initiate the precipitation of CaC₂O₄ under these conditions.

Check Your Learning

If a solution contains 0.0020 mol of ${\text{CrO}}_4{}^{\text{2−}}$ per liter, what concentration of Ag⁺ ion must be reached by adding solid AgNO₃ before Ag₂CrO₄ begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.

Answer:

4.5 $\times$ 10^–9 M

It is sometimes useful to know the concentration of an ion that remains in solution after 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 type as that in Example 15.9—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.

Due to their light sensitivity, mixtures of silver halides are used in fiber optics 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 film. Even though AgCl (K_sp = 1.6 $\times$ 10^–10), AgBr (K_sp = 5.0 $\times$ 10^–13), and AgI (K_sp = 1.5 $\times$ 10^–16) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag⁺ to a solution of Cl^–, Br^–, and I^–; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for K_sp. However, 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 form slightly soluble compounds with the same 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 we add a precipitating agent to a solution containing both anions (or both cations). When the K_sp values of the two compounds differ by two orders of magnitude or more (e.g., 10^–2 vs. 10^–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.

Chemistry in Everyday Life

The Role of Precipitation in Wastewater Treatment

Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure 15.6). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions ${\text{(PO}}_4{}^{\text{3−}})$ are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.

Figure 15.6 Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: “eutrophication&hypoxia”/Wikimedia Commons)

One common way to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)₂. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca₅(PO4)₃OH, which then precipitates out of the solution:

$$5{\text{Ca}}^{\text{2+}}+{\text{3PO}}_4{}^{\text{3−}}+{\text{OH}}^{\text{−}}\rightleftharpoons {\text{Ca}}_5{{\text{(PO}}_4)}_{\text{3}}\text{·}\text{OH}(s)$$

15.46

The precipitate is then 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(III) chloride and aluminum sulfate.

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

Selective precipitation can also be used in qualitative analysis. In this method, reagents are added to an unknown chemical mixture in order to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the addition of the reagent can be used to determine whether the ion is present in the solution.

Link to Learning

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

Common Ion Effect

As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH₃CO₂, 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 ${\text{H}}_3{\text{O}}^{\text{+}}$ to compensate for the increased acetate ion concentration. This increases the concentration of CH₃CO₂H:

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

The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:

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

Link to Learning

View this simulation to see how the common ion effect works with different concentrations of salts.

Example 15.12

Common Ion Effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr₂). The K_sp of CdS is 1.0 $\times$ 10^–28.

Solution

The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr₂ concentration as a contributor of cadmium ions:

$$\text{CdS}(s)\rightleftharpoons {\text{Cd}}^{\text{2+}}(aq)+{\text{S}}^{\text{2−}}(aq)$$

15.55

$$K_{\text{sp}}=[{\text{Cd}}^{\text{2+}}][{\text{S}}^{\text{2−}}]=1.0\times {10}^{-28}$$

15.56

$$(0.010+x)(x)=\text{1.0}\times {10}^{-28}$$

15.57

$$x^2+\text{0.010}x-\text{1.0}\times {10}^{-28}=0$$

15.58

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the K_sp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our K_sp expression, we would now get:

$$K_{\text{sp}}=[{\text{Cd}}^{\text{2+}}]{[\text{S}}^{\text{2−}}]=\text{1.0}\times {10}^{-28}$$

15.59

$$(0.010)(x)=\text{1.0}\times {10}^{-28}$$

15.60

$$x=\text{1.0}\times {10}^{-26}$$

15.61

Therefore, the molar solubility of CdS in this solution is 1.0 $\times$ 10^–26 M.

Check Your Learning

Calculate the molar solubility of aluminum hydroxide, Al(OH)₃, in a 0.015-M solution of aluminum nitrate, Al(NO₃)₃. The K_sp of Al(OH)₃ is 2 $\times$ 10^–32.

Answer:

4 $\times$ 10^–11 M