Question

calculate the concentration of $s^{2-}$ in an aqueous solution of 0.107 m hydrosulfuric acid, $h_2s$ (aq).
$s^{2-}$ = m.
submit answer retry entire group 6 more group attempts remaining

Explanation:

Step1: Write the dissociation equations

$$\ce{H2S(aq) <=> H+(aq) + HS-(aq)}$$
$$K_{a1}=\frac{[\ce{H+}][\ce{HS-}]}{[\ce{H2S}]} = 1.3\times10^{-7}$$
$$\ce{HS-(aq) <=> H+(aq) + S^{2-}(aq)}$$
$$K_{a2}=\frac{[\ce{H+}][\ce{S^{2-}}]}{[\ce{HS-}]} = 7.1\times10^{-15}$$

Step2: Assume first - step dissociation dominates

Since \(K_{a1}\gg K_{a2}\), assume \([\ce{H+}]\approx[\ce{HS-}]\) from the first - step dissociation. Let \(x = [\ce{H+}]=[\ce{HS-}]\) and \([\ce{H2S}]\approx0.107\ M\) (because \(x\) is very small compared to \(0.107\) due to small \(K_{a1}\)).
Using \(K_{a1}=\frac{[\ce{H+}][\ce{HS-}]}{[\ce{H2S}]}\), we have \(1.3\times 10^{-7}=\frac{x\cdot x}{0.107}\)
Solving for \(x\): \(x^{2}=1.3\times10^{-7}\times0.107\), \(x = [\ce{H+}]=[\ce{HS-}]\approx\sqrt{1.3\times10^{-7}\times0.107}\approx1.18\times10^{-4}\ M\)

Step3: Use \(K_{a2}\) to find \([\ce{S^{2 -}}]\)

From \(K_{a2}=\frac{[\ce{H+}][\ce{S^{2-}}]}{[\ce{HS-}]}\), and since \([\ce{H+}]\approx[\ce{HS-}]\) (from step 2), then \([\ce{S^{2-}}]=K_{a2}\)

Answer:

\(7.1\times 10^{-15}\)