8.11 - pH and Solubility

Overview

Some salts dissolve much better in acid than in water, and some don't care about pH at all. The difference is whether one of the salt's ions reacts with

\mathrm{H_3O^{+}}

\mathrm{OH^{-}}

. If it does, changing the pH pulls that ion out of solution, and Le Châtelier's principle says more solid dissolves to replace it. This topic is qualitative: the AP exam asks you to predict and explain the direction of the change, not to calculate solubility at a given pH.

The idea: remove an ion, and more solid dissolves

A slightly soluble salt in water reaches an equilibrium with its ions (Topic 7.11). For calcium fluoride:

\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)}

K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2

Anything that lowers the concentration of one of the ions makes $Q < K_{sp}$ , so more solid dissolves until $Q = K_{sp}$ again. The salt is more soluble.

Anything that raises the concentration of one of the ions (a common ion, Topic 7.12) makes $Q > K_{sp}$ , so solid forms. The salt is less soluble.

K_{sp}

itself doesn't change with pH; only temperature changes it. What changes is the solubility, how much solid has to dissolve before the ion concentrations satisfy

K_{sp}

Three kinds of ions make a salt's solubility depend on pH: the hydroxide ion, an ion that's a weak base, and an ion that's a weak acid.

Salts that contain hydroxide

Metal hydroxides such as

\mathrm{Mg(OH)_2}

and

\mathrm{Fe(OH)_3}

release

\mathrm{OH^{-}}

when they dissolve:

\mathrm{Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq)}

Add acid (lower the pH): $\mathrm{H_3O^{+}}$ reacts with $\mathrm{OH^{-}}$ , which lowers $\mathrm{[OH^{-}]}$ . Now $Q < K_{sp}$ , so more $\mathrm{Mg(OH)_2}$ dissolves. The overall reaction is:

\mathrm{Mg(OH)_2(s) + 2H_3O^{+}(aq) \rightarrow Mg^{2+}(aq) + 4H_2O(l)}

Add base (raise the pH): extra $\mathrm{OH^{-}}$ is a common ion. $Q > K_{sp}$ , so the solubility goes down.

This is how milk of magnesia works. It's a suspension of

\mathrm{Mg(OH)_2}

, which barely dissolves in water but dissolves in stomach acid and neutralizes it.

Example 1: A saturated solution of magnesium hydroxide

A beaker holds a saturated solution of

\mathrm{Mg(OH)_2}

with extra solid on the bottom. Predict what happens to the amount of solid when each of these is added, and explain.

(a) A few drops of 6 M $\mathrm{HCl}$ . The solid decreases.

\mathrm{H_3O^{+}}

reacts with

\mathrm{OH^{-}}

, lowering

\mathrm{[OH^{-}]}

. That makes

Q = [\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2

smaller than

K_{sp}

, so more

\mathrm{Mg(OH)_2}

dissolves until

Q = K_{sp}

again.

(b) A few drops of 6 M $\mathrm{NaOH}$ . The solid increases. The added

\mathrm{OH^{-}}

makes

Q > K_{sp}

, so

\mathrm{Mg(OH)_2}

precipitates until

Q = K_{sp}

(c) Does $K_{sp}$ change in either case? No. The temperature didn't change, so

K_{sp}

stays the same. Only the solubility changes.

Salts with an anion that's a weak base

The conjugate base of a weak acid is a weak base. Anions like

\mathrm{F^{-}}

\mathrm{CO_3^{2-}}

\mathrm{PO_4^{3-}}

, and

\mathrm{CH_3COO^{-}}

all react with

\mathrm{H_3O^{+}}

. For calcium fluoride in acid:

\mathrm{F^{-}(aq) + H_3O^{+}(aq) \rightarrow HF(aq) + H_2O(l)}

This reaction removes

\mathrm{F^{-}}

, so

Q < K_{sp}

and more

\mathrm{CaF_2}

dissolves. The effect gets large once the pH drops below the p

K_a

of HF (about 3.2), because then most of the dissolved fluoride is in the HF form (Topic 8.7).

Carbonates go one step further. The carbonic acid that forms breaks down into carbon dioxide, which bubbles out of the solution:

\mathrm{CO_3^{2-}(aq) + 2H_3O^{+}(aq) \rightarrow CO_2(g) + 3H_2O(l)}

Since the

\mathrm{CO_2}

escapes, the reaction can't run backward, and a carbonate keeps dissolving as long as there's acid.

Adding base to these salts has little effect.

\mathrm{OH^{-}}

doesn't react with

\mathrm{F^{-}}

\mathrm{CO_3^{2-}}

Anions that don't react

\mathrm{Cl^{-}}

\mathrm{Br^{-}}

\mathrm{I^{-}}

, and

\mathrm{NO_3^{-}}

are conjugate bases of strong acids. They're such weak bases that they don't react with

\mathrm{H_3O^{+}}

(Topic 8.6). So adding acid doesn't change the solubility of salts like

\mathrm{AgCl}

\mathrm{AgBr}

, or

\mathrm{PbI_2}

Example 2: Telling two silver salts apart (AP-style)

A student has two unlabeled powders. One is

\mathrm{AgCl}

and the other is

\mathrm{Ag_2CO_3}

. She adds 6 M

\mathrm{HNO_3}

to a small sample of each. Sample 1 fizzes and the solid dissolves. Sample 2 shows no visible change.

(a) Identify each sample and justify your answer with an equation. Sample 1 is

\mathrm{Ag_2CO_3}

. Carbonate is a weak base, so it reacts with the acid and leaves as

\mathrm{CO_2}

gas (the fizzing):

\mathrm{Ag_2CO_3(s) + 2H_3O^{+}(aq) \rightarrow 2Ag^{+}(aq) + CO_2(g) + 3H_2O(l)}

Sample 2 is

\mathrm{AgCl}

\mathrm{Cl^{-}}

is the conjugate base of a strong acid and doesn't react with

\mathrm{H_3O^{+}}

, so the acid doesn't change its solubility.

(b) Why is $\mathrm{HCl}$ a poor choice of acid for this test?

\mathrm{HCl}

adds

\mathrm{Cl^{-}}

. Sample 1 would still fizz, but the

\mathrm{Ag^{+}}

it released would immediately precipitate as

\mathrm{AgCl}

, so both test tubes would end up with a solid in them.

\mathrm{HNO_3}

avoids this because all nitrate salts are soluble.

Salts with an ion that's a weak acid

If one of the salt's ions can donate a proton, raising the pH removes it:

\mathrm{OH^{-}}

turns it into its conjugate base. So these salts become more soluble as the pH goes up. Cream of tartar, potassium hydrogen tartrate (

\mathrm{KHC_4H_4O_6}

), is only slightly soluble in water. Its anion is a weak acid:

\mathrm{KHC_4H_4O_6(s) \rightleftharpoons K^{+}(aq) + HC_4H_4O_6^{-}(aq)}

\mathrm{HC_4H_4O_6^{-}(aq) + OH^{-}(aq) \rightarrow C_4H_4O_6^{2-}(aq) + H_2O(l)}

Adding base removes

\mathrm{HC_4H_4O_6^{-}}

, so

Q < K_{sp}

and more solid dissolves. Nothing new precipitates, since potassium salts are soluble. (

\mathrm{HC_4H_4O_6^{-}}

is amphiprotic, so a strongly acidic solution also dissolves more of it, by turning it into tartaric acid,

\mathrm{H_2C_4H_4O_6}

Summary

Ion in the salt	Examples	Adding acid (lower pH)	Adding base (higher pH)
Hydroxide, $\mathrm{OH^{-}}$	$\mathrm{Mg(OH)_2}$ , $\mathrm{Fe(OH)_3}$	more soluble	less soluble (common ion)
Anion that's a weak base	$\mathrm{CaF_2}$ , $\mathrm{CaCO_3}$ , $\mathrm{Ca_3(PO_4)_2}$	more soluble	little change
Ion that's a weak acid	$\mathrm{KHC_4H_4O_6}$	little change, unless the ion is also a weak base	more soluble
Anion of a strong acid	$\mathrm{AgCl}$ , $\mathrm{AgBr}$ , $\mathrm{PbI_2}$	no change	no change, unless the metal ion forms a hydroxide precipitate

To explain any of these on the AP exam, give three things: the reaction that removes (or adds) an ion, the effect on

Q

compared with

K_{sp}

(or the direction of the shift), and the result for solubility.

Example 3: Predicting the effect of acid

Which of these salts are more soluble in 1 M

\mathrm{HNO_3}

than in pure water:

\mathrm{CaF_2}

\mathrm{AgBr}

\mathrm{Fe(OH)_3}

\mathrm{BaCO_3}

? Justify each answer.

$\mathrm{CaF_2}$ : more soluble. $\mathrm{F^{-}}$ is a weak base. It reacts with $\mathrm{H_3O^{+}}$ to form HF, which lowers $\mathrm{[F^{-}]}$ , so $Q < K_{sp}$ and more $\mathrm{CaF_2}$ dissolves.

$\mathrm{AgBr}$ : no change. $\mathrm{Br^{-}}$ is the conjugate base of a strong acid, HBr, so it doesn't react with $\mathrm{H_3O^{+}}$ .

$\mathrm{Fe(OH)_3}$ : more soluble. $\mathrm{H_3O^{+}}$ reacts with $\mathrm{OH^{-}}$ to form water, which lowers $\mathrm{[OH^{-}]}$ , so $Q < K_{sp}$ and more dissolves.

$\mathrm{BaCO_3}$ : more soluble. $\mathrm{CO_3^{2-}}$ reacts with $\mathrm{H_3O^{+}}$ and leaves as $\mathrm{CO_2}$ gas, so the solid keeps dissolving as long as acid is present.

Reading solubility data

The graph shows the molar solubility of three salts in solutions held at different pH values. The values were calculated from

K_{sp}

and

K_a

data, and the log scale lets all three fit on one graph. You won't have to do these calculations, but you should be able to read and explain a graph like this one.

$\mathrm{AgCl}$ (dashed line): flat. $\mathrm{Cl^{-}}$ doesn't react with $\mathrm{H_3O^{+}}$ , so pH has no effect.

$\mathrm{CaF_2}$ : flat above about pH 4, then rising as the pH drops. In acid, more and more $\mathrm{F^{-}}$ becomes HF.

$\mathrm{Mg(OH)_2}$ (red line): falls steeply as the pH rises. $\mathrm{OH^{-}}$ is one of the salt's own ions, so every increase in $\mathrm{[OH^{-}]}$ pushes the dissolving equilibrium back hard. Below about pH 8.4 its solubility would be above 1 M, so the line runs off the top of the graph.

Example 4: Reading the graph

Use the solubility graph above.

(a) Roughly how many times more soluble is $\mathrm{CaF_2}$ at pH 2 than at pH 7? At pH 7 the curve sits at about

2 \times 10^{-4}

M. At pH 2 it's a little above

10^{-3}

M. So

\mathrm{CaF_2}

is roughly 7 times more soluble at pH 2.

(b) Explain why $\mathrm{CaF_2}$ 's solubility doesn't change between pH 5 and pH 11. In that range the pH is well above the p

K_a

of HF, so almost all the dissolved fluoride stays as

\mathrm{F^{-}}

. Nothing removes it, so the solubility is the same as in pure water.

(c) A student wants to dissolve as much $\mathrm{Mg(OH)_2}$ as possible. Should she use a solution at pH 5 or pH 10? Explain. pH 5. The lower

\mathrm{[OH^{-}]}

at pH 5 keeps

Q

below

K_{sp}

until far more

\mathrm{Mg(OH)_2}

has dissolved. At pH 10, the high

\mathrm{[OH^{-}]}

acts as a common ion and keeps the solubility low.

Where you see this

Teeth. Tooth enamel is mostly hydroxyapatite, $\mathrm{Ca_5(PO_4)_3OH}$ . Both $\mathrm{OH^{-}}$ and $\mathrm{PO_4^{3-}}$ react with acid, so the acids that mouth bacteria make from sugar slowly dissolve enamel. Fluoride replaces some $\mathrm{OH^{-}}$ to form fluorapatite, $\mathrm{Ca_5(PO_4)_3F}$ . $\mathrm{F^{-}}$ is a much weaker base than $\mathrm{OH^{-}}$ , so fluorapatite holds up better in acid.

Stone and shells. Limestone, marble, and seashells are mostly $\mathrm{CaCO_3}$ , which dissolves in acid. Acid rain slowly erodes marble statues, and more acidic seawater makes it harder for shellfish to build their shells.

Common mistakes

Saying acid makes every salt more soluble. It only matters if one of the ions reacts with $\mathrm{H_3O^{+}}$ : hydroxide or the conjugate base of a weak acid. $\mathrm{AgCl}$ doesn't care.

Saying $K_{sp}$ changes with pH. $K_{sp}$ is constant at a given temperature. The solubility changes.

Writing "the acid dissolves it" and stopping. Name the ion that reacts, write the reaction, then explain the shift ( $Q < K_{sp}$ , so more solid dissolves).

Getting hydroxides backward in base. Adding $\mathrm{OH^{-}}$ to a metal hydroxide makes it less soluble (common-ion effect).

Testing silver or lead salts with $\mathrm{HCl}$ . The $\mathrm{Cl^{-}}$ precipitates the metal ion. Use $\mathrm{HNO_3}$ .

Calculating when you only need to explain. The AP exam doesn't ask for solubility at a given pH. It asks for the direction and the reason.

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