Exemplar Problems

Problem 8 : Calculate the pH of a solution that is 0.05 M in HNO₂ (nitrous acid) and 0.025 M in NaNO₂ (sodium nitrite). Given K_a for HNO₂ is $$(4.5 \times 10^{-4}).$$

Solution :

Given K_a for H₂S as $1.0 \times 10^{-7}$, we can set up an ICE table for the dissociation of H₂S:

Initial: $$[H_2S] = 0.1 M$$ $$[H^+] = 0\ \text{M}$$ $$[HS^-] = 0 M$$

Change: $$[H_2S] decreases by (-x)$$ $$[H^+] increases by (x)$$ $$[HS^-] increases by (x)$$

Equilibrium: $$[H_2S] - (x) M$$ $$[H^+] = x\ \text{M}$$ $$[HS^-] + (x) M$$

Using the K_a expression: $$[K_a = \frac{[H^+][HS^-]}{[H_2S]} = \frac{(x)(x)}{0.1 - x}]$$

Since (x) is small compared to 0.1, we can approximate (0.1 - x) as (0.1):

$$[1.0 \times 10^{-7} = \frac{x^2}{0.1}]$$

Solve for (x): $$[x^2 = 1.0 \times 10^{-7} \times 0.1]$$ $$[x^2 = 1.0 \times 10^{-8}]$$ $$[x = \sqrt{1.0 \times 10^{-8}}]$$ $$[x \approx 1.0 \times 10^{-4}]$$

Now, we can calculate the H⁺ concentration and the pH: $$[H^+] = x \approx 1.0 \times 10^{-4}] M$$ $$[pH = -\log[H^+] = -\log(1.0 \times 10^{-4})]$$ $$[pH = 4]$$

So, the pH of the solution is 4.