What is the pH of a 1.0 M solution of Na₂SO₃? Ka₁ = 1.3 × 10^-2, Ka₂ = 6.2 × 10^-8 ? A) 6.8 B) 7.2 C) 7.0 D) 10.4 E) 3.6

Question

Quizplus · Accepted Answer

The given compound is a long and complex molecule, but we can simplify it to NaX, where X represents the polyatomic anion ₂SO₃. The acid-base reaction for NaX in water is: NaX + H2O ⇌ HX + NaOH where HX is the acidic form of X. Since X has two acidic hydrogens, it can donate two protons to the solution. Therefore, we need to consider two equilibria: HX ⇌ H+ + X- X- ⇌ H+ + X₂SO₃- The corresponding acid dissociation constants are: Ka1 = [H+][X-]/[HX] = 1.3 × 10^-2 Ka2 = [H+][X₂SO₃-]/[X-] = 6.2 × 10^-8 At equilibrium, the concentration of NaOH is equal to that of HX, which is x. The concentration of NaX is 1.0 M (assuming complete dissolution). Therefore, at equilibrium: [NaX] = [HX] = x [NaOH] = x Using the Ka1 expression, we can write: Ka1 = x^2/(1.0 - x) = 1.3 × 10^-2 Solving for x gives: x = 2.95 × 10^-2 This is the concentration of H+ in the solution, which we can convert to pH: pH = -log[H+] = 10.53 This value is closer to answer choice D (10.4) than any other option.

What Is the PH of a 1