Exemplar Problems
Problem 10 : Calculate the pH of a solution that is 0.1 M in H2S (hydrogen sulfide) and 0.05 M in (NaHS) (sodium hydrosulfide). Given Ka for H2S is $$(1.2 \times 10^{-7}).$$
Solution :
To calculate the pH of the solution, we need to consider the ionization of H2S (hydrogen sulfide) and (NaHS) (sodium hydrosulfide) in water.
The ionization reactions are as follows:
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Ionization of H2S: $$[H_2S \rightleftharpoons H^+ + HS^-]$$
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Ionization of (NaHS): $$[NaHS \rightleftharpoons Na^+ + HS^-]$$
Given that Ka for H2S is $$(1.2 \times 10^{-7}),$$ we can write the equilibrium expression for the ionization of H2S:
$$[K_a = \frac{[H^+][HS^-]}{[H_2S]}]$$
We can use an ICE (initial, change, equilibrium) table to find the concentrations of H+ and HS- ions at equilibrium.
For the ionization 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:
$$[1.2 \times 10^{-7} = \frac{(x)(x)}{0.1 - x}]$$
Since (x) is small compared to 0.1, we can approximate (0.1 - x) as 0.1:
$$[1.2 \times 10^{-7} = \frac{x^2}{0.1}]$$
Now, solve for (x):
$$[x^2 = (1.2 \times 10^{-7})(0.1)]$$ $$[x^2 = 1.2 \times 10^{-8}]$$ $$[x = \sqrt{1.2 \times 10^{-8}}]$$ $$[x \approx 1.10 \times 10^{-4}]$$
Now, for the ionization of (NaHS):
Since (NaHS) is a salt, it dissociates completely into its ions. So, the concentration of HS- ions from (NaHS) is 0.05 M.
Now, we can calculate the total concentration of HS- ions as the sum of the concentrations from H2S and (NaHS):
Total $$[HS^-] = 0.05 M (from (NaHS)) + 1.10 \times 10^{-4} M (from (H_2S)) ≈ 0.05 M$$
To find the pH, we first calculate the (pH) using the H+ ion concentration:
$$[pH = -\log[H^+]]$$
Since (H^+) ions come from the ionization of both H2S and (NaHS), we can calculate the H+ ion concentration:
$$[H^+] = x (from H2S) + 0 (from (NaHS)) ≈ 1.10 \times 10^{-4} M$$
Now, calculate the pH:
$$[pH = -\log(1.10 \times 10^{-4})]$$
Calculate (pH):
$$[pH \approx 3.96]$$
So, the pH of the solution is approximately 3.96.
BETA
