What is the pH of a 0.10 M solution of H₂SO₃, given that its pKₐ is 1.77 and the change in concentration is small?

• A. 4.77
• B. 2.07
• C. 1.77
• D. 3.77

Answer: (B)

Verified step by step guidance

1. Start by understanding that H₂SO₃ is a weak acid, and its dissociation in water can be represented by the equation: ${H₂\mathrm{SO}₃}_2(aq)\rightleftharpoons H_2(aq)+{\mathrm{HSO}₃}_2(aq)$.
2. Use the given pKₐ value to find the Kₐ using the formula: $Kₐ={10}^{-\mathrm{pK}ₐ}$. Substitute pKₐ = 1.77 into the equation to find Kₐ.
3. Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium. Initially, the concentration of H₂SO₃ is 0.10 M, and the concentrations of H⁺ and HSO₃⁻ are 0 M.
4. Assume that the change in concentration of H₂SO₃ is small, so the equilibrium concentration of H₂SO₃ is approximately 0.10 M. Let x be the concentration of H⁺ and HSO₃⁻ at equilibrium. The equilibrium expression is: $Kₐ=\frac{x^2}{0.10}$.
5. Solve for x, which represents the concentration of H⁺ ions. Once x is found, calculate the pH using the formula: $\mathrm{pH}=-{10}^{-x}$.