# 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:

• ${K}_{\text{a}}={K}_{\text{b}}\phantom{\rule{0.2778em}{0ex}}⟹\phantom{\rule{0.2778em}{0ex}}$ neutral
• ${K}_{\text{a}}>{K}_{\text{b}}\phantom{\rule{0.2778em}{0ex}}⟹\phantom{\rule{0.2778em}{0ex}}$ acidic
• ${K}_{\text{a}}<{K}_{\text{b}}\phantom{\rule{0.2778em}{0ex}}⟹\phantom{\rule{0.2778em}{0ex}}$ 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:

CO2(g) + H2O(l) → H2CO3(aq)

H2CO3(aq) + H2O(l) → HCO3(aq) + H3O+(aq).

## Example

Will (NH4)2SO4(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)2SO4(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) ⇌ NH3(aq) + H3O+(aq), ${K}_{\text{a}}=5.8×{10}^{-10}\text{.}$

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 H2SO4(aq), which is the same as ${K}_{\text{a}}$ for HSO4(aq). We can use this to find the ionization constant for the conjugate base of HSO4(aq), which is SO42−(aq). Since ${K}_{\text{a}}{K}_{\text{b}}={K}_{\text{w}}\text{,}$

$Kb=KwKa2=1.0×10−141.0×10−2=1.0×10−12.$

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.