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Chapter 14: Problem 107

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) solution. Assume \(K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\)

Short Answer

Expert verified
In summary, for a 0.10 M H₂S solution with Ka₁ = 1.0 x 10⁻⁷ and Ka₂ = 1.0 x 10⁻¹⁹, the pH is 4, and the concentration of S²⁻ ions is 1.0 x 10⁻¹⁹ M.

Step by step solution

01

Write the ionization reactions and dissociation constants

For the two-step ionization of H₂S, we have the following reactions and dissociation constants: 1. First ionization: \(H_2S \rightleftharpoons H^+ + HS^-\) Dissociation constant Ka₁: \[K_{a_{1}} = \frac{[H^+][HS^-]}{[H_{2}S]} = 1.0×10^{-7}\] 2. Second ionization: \(HS^- \rightleftharpoons H^+ + S^{2-}\) Dissociation constant Ka₂: \[K_{a_{2}} = \frac{[H^+][S^{2-}]}{[HS^-]} = 1.0×10^{-19}\]
02

Write the ionization expressions

Let x mol of H₂S ionizes. So, the concentration of each species after the first ionization can be expressed as follows: \[ [H^+] = x \] \[ [HS^-] = x \] \[ [H_2S] = 0.10 - x \] Now, substitute these expressions in Ka₁: \[1.0×10^{-7} = \frac{x \cdot x}{0.10 - x}\]
03

Solve for x and calculate the pH

As Ka₁ is a weak acid dissociation constant, the value of x will be significantly smaller than the initial concentration (0.10) of H₂S. Therefore, we can assume \(x ≪ 0.10\). This allows us to simplify the equation for x: \[1.0×10^{-7} ≈ \frac{x^2}{0.10}\] Solve for x: \[x ≈ \sqrt{1.0×10^{-7} \cdot 0.10} = 1.0×10^{-4}\] So, the concentration of H⁺ ions is approximately 1.0 x 10⁻⁴ M. Now, the pH can be calculated: \[pH = -\log{[H^+]} = -\log{(1.0×10^{-4})} = 4\]
04

Calculate the concentration of S²⁻ using Ka₂

Knowing the concentration of the H⁺ ion and the first ionization product (HS⁻), we can now proceed to calculate the concentration of the sulfide ion (S²⁻) using Ka₂: \[1.0×10^{-19} = \frac{(1.0×10^{-4}) [S^{2-}]}{x}\] Substitute x = 1.0 x 10⁻⁴ \[1.0×10^{-19} = \frac{(1.0×10^{-4}) [S^{2-}]}{1.0×10^{-4}}\] Solve for the concentration of S²⁻: \[[S^{2-}] = 1.0×10^{-19}\] So, the concentration of sulfide ions (S²⁻) is 1.0 x 10⁻¹⁹ M. To summarize, the pH of the 0.10 M H₂S solution is 4, and the concentration of S²⁻ ions is 1.0 x 10⁻¹⁹ M.

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