Exemplar Problems

Problem 2 : A weak acid (HA) with a dissociation constant K_a of $$(1.8 \times 10^{-5})$$ M ionizes partially in water to give H⁺ and A^- ions. Calculate the pH of a 0.05 M solution of (HA).

Solution :

Given that K_a for (HA) is $$(1.8 \times 10^{-5}),$$ we can set up an ICE (Initial, Change, Equilibrium) table for the dissociation of (HA):

Initial: $$[HA] = 0.05 M$$ $$[H^+] = 0 M$$ $$[A^-] = 0 M$$

Change: $$[HA] decreases by (-x)$$ $$[H^+] increases by (x)$$ $$[A^-] increases by (x)$$

Equilibrium: $$[HA] - (x) M$$ $$[H^+] + (x) M$$ $$[A^-] + (x) M$$

Using the K_a expression: $$[K_a = \frac{[H^+][A^-]}{[HA]}]$$

Substitute the equilibrium concentrations: $$[1.8 \times 10^{-5} = \frac{(x)(x)}{0.05 - x}]$$

Since (x) is small compared to 0.05, we can approximate $$(0.05 - x) as (0.05):$$

$$[1.8 \times 10^{-5} = \frac{x^2}{0.05}]$$

Solve for (x): $$[x^2 = 1.8 \times 10^{-5} \times 0.05]$$ $$[x^2 = 9 \times 10^{-7}]$$ $$[x = \sqrt{9 \times 10^{-7}}]$$ $$[x = 3 \times 10^{-4}]$$

Now, we can calculate the pH: $$[pH = -\log[H^+] = -\log(3 \times 10^{-4})]$$ $$[pH \approx 3.52]$$