Salt hydrolosis

A salt is an ionic compound (cation and anion). Salts are very soluble in water. When they dissolve, the ions sometimes react with water and affect the pH of the solution; we call this hydrolosis. To determine how a salt affects pH, you must look at each of its ions individually:

neutral (does not affect pH)

Cation of a strong base (e.g., Na⁺ from NaOH) or anion of a strong acid (e.g., Cl⁻ from HCl). Most of these ions are from group 1, 2, or 7, but don’t forget polyatomic ions like SO₄²⁻ from H₂SO₄.

acidic (decreases pH)

Conjugate acid of a weak base (e.g., NH₄⁺, the conjugate acid of NH₃) or metal cation with high charge density (Al³⁺ and Fe³⁺ are the main ones).

basic (increases pH)

Conjugate base of a weak acid (e.g., CH₃COO⁻, the conjugate base of CH₃COOH).

When a salt contains both a cation that decreases pH and an anion that increases pH, we must compare their ionization constants to see which one overwhelms the other:

• $K_{\text{a}}=K_{\text{b}}⟹$ neutral
• $K_{\text{a}}>K_{\text{b}}⟹$ acidic
• $K_{\text{a}}<K_{\text{b}}⟹$ basic

When a salt contains an amphoteric ion (it can donate a proton or accept one, e.g., HSO₄⁻), you must write the equation twice: once for when the ion acts as an acid and once for when it acts as a base. Then, compare the $K_{\text{a}}$ and $K_{\text{b}}$ values the same way as shown above.

There is one more special case: oxides. A metal oxide always reacts with water to form a basic solution because one of the products is the hydroxide ion. Don’t write a dissociation equation—just react the compound with water directly. For example:

CaO_(s) + H₂O_(l) → Ca²⁺_(aq) + 2 OH⁻_(aq).

A nonmetal oxide always reacts with water to form an acidic solution because one of the products is the hydronium ion. For example:

CO_2(g) + H₂O_(l) → H₂CO_3(aq)

H₂CO_3(aq) + H₂O_(l) → HCO₃⁻_(aq) + H₃O⁺_(aq).

Example

Will (NH₄)₂SO_4(s) produce a neutral, acidic, or basic solution when dissolved in water?

First, we write the dissociation reaction, indicating that the salt dissolves completely with a one-directional arrow:

(NH₄)₂SO_4(s) → 2 NH₄⁺_(aq) + SO₄²⁻_(aq).

Since NH₄⁺ is the conjugate acid of a weak base, we know that it will cause the solution to become acidic:

NH₄⁺_(aq) + H₂O_(l) ⇌ NH_3(aq) + H₃O⁺_(aq), $K_{\text{a}}=5.8\times {10}^{-10}\text{.}$

However, SO₄²⁻_(aq) will also affect the pH: it is the conjugate base of a weak acid, so it will cause the solution to become basic:

SO₄²⁻_(aq) + H₂O_(l) ⇌ HSO₄⁻_(aq) + OH⁻_(aq).

We can’t just look up the ionization constant like we did for NH₄⁺_(aq) because it isn’t on the list, at least not directly. What is on the list is $K_{\text{a2}}$ for H₂SO_4(aq), which is the same as $K_{\text{a}}$ for HSO_4(aq). We can use this to find the ionization constant for the conjugate base of HSO₄⁻_(aq), which is SO₄²⁻_(aq). Since $K_{\text{a}}{K}_{\text{b}}=K_{\text{w}}\text{,}$

$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a2}}}=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-2}}=1.0\times {10}^{-12}\text{.}$

Now we have our $K_{\text{a}}$ and $K_{\text{b}}$ values for NH₄⁺_(aq) and SO₄²⁻_(aq). They are 5.8 × 10⁻¹⁰ and 1.0 × 10⁻¹² respectively. Since $K_{\text{a}}>K_{\text{b}}\text{,}$ dissolving this salt in water will result in an acidic solution.