# 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:

• $\displaystyle K_{\text{a}} = K_{\text{b}} \Rightarrow$ neutral
• $\displaystyle K_{\text{a}} > K_{\text{b}} \Rightarrow$ acidic
• $\displaystyle K_{\text{a}} < K_{\text{b}} \Rightarrow$ 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 $\displaystyle K_{\text{a}}$ and $\displaystyle K_{\text{b}}$ 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), $\displaystyle K_{\text{a}} = 5.8 \times 10^{- 10}$.

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 $\displaystyle K_{\text{a2}}$ for H2SO4(aq), which is the same as $\displaystyle K_{\text{a}}$ for HSO4(aq). We can use this to find the ionization constant for the conjugate base of HSO4(aq), which is SO42−(aq). Since $\displaystyle K_{\text{a}} K_{\text{b}} = K_{\text{w}}$,

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

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