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 Ka for HNO2 is $$(4.5 \times 10^{-4}).$$
Solution :
Given Ka for H2S as $$(1.0 \times 10^{-7}),$$ we can set up an ICE table for the dissociation of H2S:
Initial: $$[H_2S] = 0.1 M$$ $$[H^+] = 0 M$$ $$[HS^-] = 0 M$$
Change: $$[H_2S] decreases by (-x)$$ $$[H^+] increases by (x)$$ $$[HS^-] increases by (x)$$
Equilibrium: $$[H_2S] - (x) M$$ $$[H^+] + (x) M$$ $$[HS^-] + (x) M$$
Using the Ka 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.
BETA
