Вплив pH на розчинність

We've started learning about acids and bases. But we can't forget about solubility! We used the solubility product constant to examine precipitates. But, how does pH affect solubility? Let's explore!

Solubility with Strong Conjugate Bases

Imagine we made a solution. We'll dissolve a transition metal-hydroxide. We know from our qualitative solubility rules that they have minuscule solubilities. They are practically insoluble! But, what happens if we add an acid? Hydroxide is the conjugate base of water. Water self-ionizes and is an incredibly weak acid. We learned that the conjugate of a weak acid is a strong base. We also know, acids and bases neutralize. So, in our imagined solution, the acid and hydroxide would neutralize. What direction would expect solubility to shift? Let's look at a few examples.

Recall our qualitative solubility rules for the alkaline earth-hydroxide compounds. They have low solubilities. Let's consider calcium hydroxide, $Ca(OH)_{2}$ . Its $K_{sp} is$ 6.5\times10^{-6}$. What happens if we lower the pH by adding a strong acid?

Let's start by writing the balanced chemical equations.

$Ca(OH)_{2} \rightleftharpoons Ca^{2+} + 2OH^{-}$

$H^{+} + OH^{-} \rightarrow H_{2}O$

We have two competing reactions. As the calcium hydroxide dissociates, it produces hydroxide. But, the neutralization reaction immediately consumes the hydroxide. Do you remember Le' Chatelier's principle? What happens when we reduce the product's concentration? We shift the equilibrium toward the product's side! So, more calcium hydroxide will dissolve. It will continue to dissolve until it neutralizes all the acid.

What happens if we don't have a hydroxide? Let's look at calcium carbonate, $CaCO_{3}$ . Its $K_{sp}$ is $3.3\times10^{-9}$ . We can tell from the magnitude that it is fairly insoluble. But, what happens when we add an acid?

Again, let's start by writing the balanced chemical equations.

$CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO_{3(aq)}^{2-}$

$H^{+}_{aq} + CO_{3(aq)}^{2-} \rightleftharpoons HCO_{3(aq)}^{-}$

Carbonate is the conjugate of the weak acid hydrogen carbonate. That means that carbonate is a strong base. As a side note, hydrogen carbonate is the first dissociation of the polyprotic acid, carbonic acid. So, it's amphoteric. It can act as either an acid or base. But, when it produces carbonate, it's acting as a weak acid. So, carbonate is a strong base.

By adding an acid, we consume carbonate. That shifts the dissociations of calcium carbonate to the right. So, it becomes more soluble in acidic solutions. The solubility increases for compounds with strong conjugate bases.

The dissociation of calcium carbonate is paramount in environmental chemistry. As our oceans become more acidic, calcium carbonate dissolves. Sadly, it's the reaction that is destroying coral reefs and crustaceans.

Solubility with Weak Conjugate Bases

Now, let's look at what happens when we have a weak conjugate base. Let's look at silver chloride. We know it's one of the exceptions to the solubility of the metal-halogen compounds. It's $K_{sp}$ is $1.6\times 10^{-10}$ . What happens when we add an acid?

Again, let's write the balanced chemical equations.

$AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$

$HCl_{(aq)} \rightarrow H^{+}_{aq} + Cl^{-}_{aq}$

Hydrochloric acid is a strong Arrhenius acid. So, it will fully dissociate. We can confirm this with its pKa. It's $-6.3$ ! So, when we acidify the solution, no chlorine is consumed. The equilibrium doesn't shift! So, silver chloride's solubility remains unchanged. The solubility of compounds with weak conjugate bases does not change.

Changing pH to Cause Selective Precipitation

When we studied solubility, we learned about selective precipitation. We can separate ions by manipulating precipitation reactions. Then we investigated qualitative analysis of solutions. We saw that solubility often depends on pH. So now, we can use pH changes as a tool to form precipitates.

Imagine I have a solution with Cr(II) and Cr(III) ions. Let's say the molarities are 0.001M for Cr(II) and 0.1M for Cr(III). For Cr(II)(OH_{2}), the $K_{sp}$ is $2.0\times10^{-16}$ . For Cr(III)(OH_{3}), the $K_{sp}$ is $6.3\times10^{-31}$ . What concentration of hydroxide will precipitate each ions? Which precipitates first?

First, we need to write the balanced chemical equations.

$Cr(OH)_{2(s)} \rightleftharpoons Cr^{2+}_{(aq)} + OH^{-}_{(aq)}$

$Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + OH^{-}_{(aq)}$

Now, we can use the $K_{sp}$ values to calculate the hydroxide concentration.

$K_{sp}=[Cr(II)][OH^{-}]^{2}\rightarrow [OH^{-}]=\sqrt{\frac{2\times10^{-16}}{0.001}}=4.5\times10^{-7}$

$K_{sp}=[Cr(III)][OH^{-}]^{3}\rightarrow [OH^{-}]=(\frac{6.3\times10^{-31}}{0.1})^{\frac{1}{3}}=1.8\times10^{-10}$

So, by raising the pH, we can separate Cr(II) and Cr(III) ions. The Cr(III) requires less hydroxide and will precipitate first. That's what we would expect based on the $K_{sp}$ values.

Chemistry in the World: Acid Mine Drainage

Why do we care about pH and solubility? It's paramount in environmental contexts. Let's consider acid mine drainage. It occurs when pyrite, $Fe(II)S_{2}$ , is exposed to oxygen. Typically, the pyrite is deep underground. So, very little oxygen is available. The reaction ensues over thousands of years. But, sudden oxygen exposure happens during mining or construction projects. Let's explore the reactions.

$4FeS_{2} + 14O_{2} + 4H_{2}O \rightarrow 4Fe^{2+} +8 SO_{4}^{2-} + 8H^{+}$
$4Fe^{2+}+ O_{2}+ 4H^{+} \rightarrow 4Fe^{3+} + 2H_{2}O$
$4Fe^{3+} +12H_{2}O \rightarrow 4Fe(OH)_{3(s)} + 12H^{+}$

The final reaction is the most interesting. We've produced twelve moles of acid. So, our water is very acidic. Acid mine drainage lowers the pH of creeks, streams, and rivers. The lowered pH destroys the ecosystem.

What happens if we neutralize the acid? Let's look at the solubility of iron(III) hydroxide. Its dissociation is, $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+} + 3 OH^{-}$ . It's $K_{sp}=2.8\times10^{-39}$ . So, it's not very soluble. Typically, we need a pH below 3.5 for it to dissociate. As we raise the pH, we reform iron(III) hydroxide. It has a gross yellow color. That's why it's nicknamed yellowboy. Even worse, it's toxic to fish. So, we have a problem. The acidic conditions kill fish. But, neutralizing the acid also creates a toxic environment. Acid mine drainage is a major environmental challenge.