Answer:
At equilibrium:
[HSO₄⁻] = 0.10 M;
[SO₄²⁻] = 0.037 M;
[H⁺] = 0.037 M;
All three values are in 2 sig. fig. as in Ka or the KHSO₄ concentration.
Explanation:
There's initially no SO₄²⁻ and negligible amount of H⁺ in the solution. KHSO₄ is a salt. All KHSO₄ will dissociate to form K⁺ and HSO₄⁻ ions. The initial concentration of [HSO₄⁻] will be 0.14 M.
HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. [tex]\text{K}_a = 1.3\times 10^{-2}[/tex] HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.
[tex]\begin{array}{c|ccccc}\text{R}&{\text{HSO}_4}^{-} &\rightleftharpoons& \text{H}^{+}&+&{\text{SO}_4}^{2-}\\\text{I}&0.14&&&&\\\text{C}&-x & &+x &&+x\\\text{E}&0.14 - x&&x&&x\end{array}[/tex].
[tex]\text{K}_a = 1.3 \times 10^{-2}[/tex] for [tex]{\text{HSO}_4}^{-}[/tex]. As a result,
[tex]\displaystyle \frac{[\text{H}^{+}]\cdot[{\text{SO}_4}^{2-}]}{[{\text{HSO}_4}^{-}]} = \text{K}_a[/tex].
[tex]\text{K}_a[/tex] is large. It is no longer valid to approximate that [tex][{\text{HSO}_4}^{-}][/tex] at equilibrium is the same as its initial value.
[tex]\displaystyle \frac{x^{2}}{0.14 - x} = 1.3\times 10^{-2}[/tex].
[tex]x^{2} + 1.3\times 10^{-2}\;x - 0.14\times 1.3\times 10^{-2} = 0[/tex].
Solve the quadratic equation for [tex]x[/tex]. [tex]x\ge 0[/tex] since [tex]x[/tex] represents a concentration.
[tex]x = 0.0366538[/tex].
The least significant number in the question comes with 2 sig. fig. Round the results to 2 sig. fig:
