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Chapter 14: 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}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}$

Short Answer

Expert verified
In a 0.10 M H₂S solution with given ionization constants \(K_{a_{1}} = 1.0 \times 10^{-7}\) and \(K_{a_{2}} = 1.0 \times 10^{-19}\), the pH is calculated to be 4 and the concentration of S²⁻ ions is \(1.0 \times 10^{-15}\) M.

Step by step solution

01

Write Equilibrium Equations

For ionization constants, we need to set up the equilibrium expressions for H₂S: Equation 1: \(H_{2}S \rightleftharpoons H^{+} + HS^{-}\) with equilibrium constant, Ka₁ Equation 2: \(HS^{-} \rightleftharpoons H^{+} + S^{2-}\) with equilibrium constant, Ka₂
02

Write Expressions for Ka₁ and Ka₂

Now we will write the expressions for the ionization constants Ka₁ and Ka₂: \(K_{a_{1}} = \frac{[H^{+}][HS^{-}]}{[H_{2}S]}\) and \(K_{a_{2}} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]}\)
03

Approximate the Concentrations

Since Ka₂ is much smaller than Ka₁, we can assume for the time being that the majority of hydrogen ions come from the first ionization: Let x be the concentration of H⁺ ions, so [H⁺] = x. For the first equilibrium, [HS⁻] ≈ x and [H₂S] ≈ 0.10 M. We can ignore the contribution from the second ionization in this step.
04

Solve for x (H⁺ Concentration)

Using the expression for Ka₁ and the given value of Ka₁, we can solve for x: \(1.0 \times 10^{-7} = \frac{x^2}{0.10}\) \(x^2 = 1.0 \times 10^{-8}\) \(x = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4}\) Now we have the H⁺ ion concentration, and we can calculate the pH.
05

Calculate the pH

Using the H⁺ ion concentration, we can determine the pH. pH = -log10([H⁺]) = -log10(1.0 × 10⁻⁴) = 4 Now, we will find the concentration of S²⁻ ions.
06

Solve for [S²⁻] Concentration

Using the second equilibrium equation and the value of Ka₂, we can find the concentration of S²⁻ ions, [S²⁻], by substituting [H⁺]=1.0 x 10⁻⁴ and assuming that [HS⁻] ≈ x which we found in the previous steps: \(K_{a_{2}} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]}\) \(1.0 \times 10^{-19} = \frac{(1.0 \times 10^{-4})[S^{2-}]}{1.0 \times 10^{-4}}\) \[S^{2-} = 1.0 \times 10^{-15}\] So, the pH of the 0.10 M H₂S solution is 4 and the concentration of S²⁻ ions is \(1.0 \times 10^{-15}\) M.

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Most popular questions from this chapter

A typical vitamin \(\mathrm{C}\) tablet (containing pure ascorbic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\) ) weighs $500 . \mathrm{mg}\( . One vitamin \)\mathrm{C}$ tablet is dissolved in enough water to make 200.0 \(\mathrm{mL}\) of solution. Calculate the \(\mathrm{pH}\) of this solution. Ascorbic acid is a diprotic acid.

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