Question

What are the concentrations of hso4−, so42−, and h+ in a 0.14 m khso4 solution? (hint: h2so4 is a strong acid; ka for hso4− = 1.3 10-2.) hso4−?

Answer 1

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. $\text{K}_a = 1.3\times 10^{-2}$ 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⁻.

$\begin{array}{c|ccccc}\text{R}&{\text{HSO}_4}^{-} &\rightleftharpoons& \text{H}^{+}&+&{\text{SO}_4}^{2-}\\\text{I}&0.14&&&&\\\text{C}&-x & &+x &&+x\\\text{E}&0.14 - x&&x&&x\end{array}$.

$\text{K}_a = 1.3 \times 10^{-2}$ for ${\text{HSO}_4}^{-}$. As a result,

$\displaystyle \frac{[\text{H}^{+}]\cdot[{\text{SO}_4}^{2-}]}{[{\text{HSO}_4}^{-}]} = \text{K}_a$.

$\text{K}_a$ is large. It is no longer valid to approximate that $[{\text{HSO}_4}^{-}]$ at equilibrium is the same as its initial value.

$\displaystyle \frac{x^{2}}{0.14 - x} = 1.3\times 10^{-2}$.

$x^{2} + 1.3\times 10^{-2}\;x - 0.14\times 1.3\times 10^{-2} = 0$.

Solve the quadratic equation for $x$. $x\ge 0$ since $x$ represents a concentration.

$x = 0.0366538$.

The least significant number in the question comes with 2 sig. fig. Round the results to 2 sig. fig:

• $[{\text{SO}_4}^{2-}] = x = 0.037\;\text{mol}\cdot\text{L}^{-1}$;
• $[{\text{H}^{+}] = x = 0.037\;\text{mol}\cdot\text{L}^{-1}$;
• $[{\text{HSO}_{4}^{-}] = 0.14 - x = 0.10\;\text{mol}\cdot\text{L}^{-1}$.