An aqueous solution at 25.0°C contains [H⁺] = 0.080 M.What is the pH of the solution? A)1.10 B)-1.10 C)13.0 D)0.0800 E)1.30 × 10^-1³

Question

Quizplus · Accepted Answer

The given concentration is of the acid H2S, which can be written as HS- in water. The dissociation of HS- in water can be written as follows:

HS- + H2O ↔ H3O+ + S2-

The equilibrium constant for this reaction is given as Ka = [H3O+][S2-]/[HS-]. Assuming the concentration of H3O+ and S2- to be negligible, and [HS-] = 0.080 M, the concentration of [H3O+] can be calculated as follows:

Ka = [H3O+][S2-]/[HS-]
10^-7.2 = [H3O+][10^-7.2]/[0.080]
[H3O+] = 10^-7.2 × 0.080
[H3O+] = 5.01 × 10^-9 M

Therefore, pH = -log[H3O+] = -log(5.01 × 10^-9) ≈ 1.10.

The answer is A)1.10.

An Aqueous Solution at 25