14.4 Hydrolysis of salt solutions  (Page 4/5)

Determining the acidic or basic nature of salts

(a) KBr

(b) NaHCO ₃

(c) NH ₄ Cl

(d) Na ₂ HPO ₄

(e) NH ₄ F

Solution

(a) The K ⁺ cation and the Br ⁻ anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.

(b) The Na ⁺ cation is a spectator, and will not affect the pH of the solution; while the ${\text{HCO}}_3{}^{\text{−}}$ anion is amphiprotic, it could either behave as an acid or a base. The K _a of ${\text{HCO}}_3{}^{\text{−}}$ is 5.6 $\times$ 10 ⁻¹¹ , so the K _b of its conjugate base is $\frac{1.0\times {10}^{\mathrm{-14}}}{5.6\times {10}^{\mathrm{-11}}}=1.8\times {10}^{\mathrm{-4}}.$

Since K _b >> K _a , the solution is basic.

(c) The ${\text{NH}}_4{}^{\text{+}}$ ion is acidic and the Cl ⁻ ion is a spectator. The solution will be acidic.

(d) The Na ⁺ ion is a spectator, while the ${\text{HPO}}_4{}^{\text{2−}}$ ion is amphiprotic, with a K _a of 4.2 $\times$ 10 ⁻¹³

so that the K _b of its conjugate base is $\frac{1.0\times {10}^{\mathrm{-14}}}{4.2\times {10}^{\mathrm{-13}}}=2.4\times {10}^{\mathrm{-2}}.$ Because K _b >> K _a , the solution is basic.

(e) The ${\text{NH}}_4{}^{\text{+}}$ ion is listed as being acidic, and the F ⁻ ion is listed as a base, so we must directly compare the K _a and the K _b of the two ions. K _a of ${\text{NH}}_4{}^{\text{+}}$ is 5.6 $\times$ 10 ⁻¹⁰ , which seems very small, yet the K _b of F ⁻ is 1.4 $\times$ 10 ⁻¹¹ , so the solution is acidic, since K _a > K _b .

Check your learning

(a) K ₂ CO ₃

(b) CaCl ₂

(c) KH ₂ PO ₄

(d) (NH ₄ ) ₂ CO ₃

(e) AlBr ₃