Chapter 13: Problem 109

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10-M \mathrm{H}_{2} \mathrm{S}\) solution. Assume \(K_{\mathrm{a}_{\mathrm{t}}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\).

Short Answer

Expert verified

The pH of the 0.10 M H₂S solution is 4, and the concentration of S²⁻ ions is \(1.0\times10^{-19} M\).

Step by step solution

Write down the major equilibrium reaction and its dissociation constant.

We begin by analyzing the major reactant, \( H_2S\). It can lose two protons sequentially: \[H_{2}S \rightleftharpoons H^+ + HS^-\] and \[HS^- \rightleftharpoons H^+ + S^{2-}\] However, given the significant difference in magnitudes of the dissociation constants (\(K_{a1} = 1.0\times10^{-7}\) and \(K_{a2} = 1.0\times10^{-19}\)), we will primarily focus on the first dissociation since its contribution to pH will be dominant while estimating \( [S^{2-}]\). So, for the first reaction, \[K_{a1} = \frac{[H^+][HS^-]}{[H_{2}S]}\]

Set up an equilibrium expression for [H⁺] ions.

Initial conc. of H₂S is 0.1 M and of H⁺, HS⁻ is zero in the start. As the reaction reaches equilibrium, some of the H₂S will dissociate to form H⁺ and HS⁻ ions. Since the reaction ratio is 1:1:1, we represent concentrations as follows: [H⁺] = x [HS⁻] = x [H₂S] = 0.1 - x Then, \[ K_{a1} = \frac{x^2}{0.1 - x} \Longrightarrow 1.0\times10^{-7} = \frac{x^2}{0.1 - x}\]

Solve for [H⁺] and find pH.

To solve for x, we can make an approximation since \(K_{a1}\) value is low enough. We can assume that x << 0.1, which means we can approximate the denominator with simply 0.1: \[1.0\times10^{-7} = \frac{x^2}{0.1}\] \[x^2 = 1.0\times10^{-8}\] \[x = 1.0\times10^{-4}\] Therefore, [H⁺] = \(1.0\times10^{-4} M\), and to find the pH, we use the formula: \[pH = -\log([H^+]) = -\log(1.0\times10^{-4}) = 4\] Thus, the pH of the H₂S solution is 4.

Find the concentration of [S²⁻] by considering the second dissociation.

Since the second dissociation is a minor one with a very low dissociation constant (K₂), we can assume that the concentration of HS⁻ ions is approximately equal to the [H⁺] calculated before: [HS⁻] = \(1.0\times10^{-4} M\) Now, for the second equilibrium, we have: \[K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}\] Substitute the given values into the equation: \[1.0\times10^{-19} = \frac{(1.0\times10^{-4})([S^{2-}])}{(1.0\times10^{-4})}\] Solve for [S²⁻]: \[[S^{2-}] = 1.0\times10^{-19}\] So, the concentration of S²⁻ ions in the 0.1 M H₂S solution is \(1.0\times10^{-19} M\).

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Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10-M \mathrm{H}_{2} \mathrm{S}\) solution. Assume \(K_{\mathrm{a}_{\mathrm{t}}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\).

Short Answer

Step by step solution

Write down the major equilibrium reaction and its dissociation constant.

Set up an equilibrium expression for [H⁺] ions.

Solve for [H⁺] and find pH.

Find the concentration of [S²⁻] by considering the second dissociation.

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