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 K_a 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:

1. Ionization of H₂S: $$[H_2S \rightleftharpoons H^+ + HS^-]$$

2. Ionization of (NaHS): $$[NaHS \rightleftharpoons Na^+ + HS^-]$$

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

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:

$$[1.2 \times 10^{-7} = \frac{(x)(x)}{0.1 - x}]$$

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.