Pregunta
Calculating the pH of a salt solution Calculate the pH at \( 25^{\circ} \mathrm{C} \) of a 0.34 M solution of sodium benzoate \( \left(\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right) \). Note that benzoic acid \( \left(\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right) \) is a weak acid with a \( p K_{a} \) of 4.20 . Round your answer to 1 decimal place.
Ask by Peters Sherman. in the United States
Mar 17,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
The pH of the 0.34 M sodium benzoate solution at \( 25^{\circ} \mathrm{C} \) is approximately 8.9.
Solución
We start by writing the hydrolysis reaction for the benzoate ion in water:
\[
\mathrm{C}_6\mathrm{H}_5\mathrm{CO}_2^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HC}_6\mathrm{H}_5\mathrm{CO}_2 + \mathrm{OH}^-
\]
Because benzoic acid is weak, the benzoate ion acts as a weak base with a base ionization constant \( K_b \). The relationship between the acid and base constants is:
\[
K_a \cdot K_b = K_w
\]
Given \( K_a = 10^{-4.20} \) and \( K_w = 1.0 \times 10^{-14} \), we calculate \( K_b \):
\[
K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.20}} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-5}} \approx 1.58 \times 10^{-10}
\]
Next, let \( x \) be the concentration of \(\mathrm{OH}^-\) produced (in M). We assume the initial concentration of benzoate is 0.34 M and that the change \( x \) is small relative to 0.34 M. The equilibrium expression becomes:
\[
K_b = \frac{x^2}{0.34}
\]
Solving for \( x \):
\[
x^2 = K_b \times 0.34 \quad \Longrightarrow \quad x = \sqrt{1.58 \times 10^{-10} \times 0.34}
\]
\[
x = \sqrt{5.37 \times 10^{-11}} \approx 7.33 \times 10^{-6}
\]
This \( x \) represents the concentration of \(\mathrm{OH}^-\) in the solution. The \( pOH \) is calculated as:
\[
pOH = -\log (7.33 \times 10^{-6}) \approx 5.1
\]
Finally, the \( pH \) of the solution is:
\[
pH = 14 - pOH = 14 - 5.1 \approx 8.9
\]
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