The Ka for benzoic acid is 6.5 × 10−5 . Calculate the pH of a 0.10 M benzoic acid solution.
The Correct Answer And Explantion is:
To calculate the pH of a 0.10 M benzoic acid (C6H5COOHC_6H_5COOHC6H5COOH) solution, we use the acid dissociation constant (KaK_aKa) and set up an equilibrium expression.
Step 1: Write the Dissociation Equation
Benzoic acid dissociates in water as: C6H5COOH⇌C6H5COO−+H+C_6H_5COOH \rightleftharpoons C_6H_5COO^- + H^+C6H5COOH⇌C6H5COO−+H+
Step 2: Set Up the ICE Table
Let the initial concentration of benzoic acid be 0.10 M. Before dissociation:
- [C6H5COOH]=0.10M[C_6H_5COOH] = 0.10 M[C6H5COOH]=0.10M
- [C6H5COO−]=0[C_6H_5COO^-] = 0[C6H5COO−]=0
- [H+]=0[H^+] = 0[H+]=0
Let xxx be the amount of benzoic acid that dissociates at equilibrium. Then, the equilibrium concentrations are:
- [C6H5COOH]=0.10−x[C_6H_5COOH] = 0.10 – x[C6H5COOH]=0.10−x
- [C6H5COO−]=x[C_6H_5COO^-] = x[C6H5COO−]=x
- [H+]=x[H^+] = x[H+]=x
Step 3: Write the KaK_aKa Expression
Ka=[C6H5COO−][H+][C6H5COOH]K_a = \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}Ka=[C6H5COOH][C6H5COO−][H+]
Substituting values: 6.5×10−5=x20.10−x6.5 \times 10^{-5} = \frac{x^2}{0.10 – x}6.5×10−5=0.10−xx2
Since KaK_aKa is small, we approximate 0.10−x≈0.100.10 – x \approx 0.100.10−x≈0.10: 6.5×10−5=x20.106.5 \times 10^{-5} = \frac{x^2}{0.10}6.5×10−5=0.10×2 x2=(6.5×10−5)×0.10x^2 = (6.5 \times 10^{-5}) \times 0.10×2=(6.5×10−5)×0.10 x2=6.5×10−6x^2 = 6.5 \times 10^{-6}x2=6.5×10−6 x=6.5×10−6=2.55×10−3x = \sqrt{6.5 \times 10^{-6}} = 2.55 \times 10^{-3}x=6.5×10−6=2.55×10−3
Thus, [H+]=2.55×10−3[H^+] = 2.55 \times 10^{-3}[H+]=2.55×10−3.
Step 4: Calculate pH
pH=−log[H+]\text{pH} = -\log [H^+]pH=−log[H+] pH=−log(2.55×10−3)\text{pH} = -\log(2.55 \times 10^{-3})pH=−log(2.55×10−3) pH≈2.59\text{pH} \approx 2.59pH≈2.59
Explanation (300 words)
The pH of a 0.10 M benzoic acid solution is calculated using the acid dissociation constant (KaK_aKa). Since benzoic acid is a weak acid, it only partially dissociates in water. We start by writing the dissociation equation and setting up an ICE table to determine the equilibrium concentrations of the species involved.
The acid dissociation constant equation is then used to solve for [H+][H^+][H+], assuming that the change in the initial concentration is negligible. This simplifies the quadratic equation to a straightforward square root calculation. By solving for xxx, which represents [H+][H^+][H+], we find it to be 2.55×10−32.55 \times 10^{-3}2.55×10−3 M.
Using the pH formula, we take the negative logarithm of the hydrogen ion concentration and find the pH to be 2.59. This result makes sense because benzoic acid is a weak acid, and a 0.10 M solution should have a pH lower than 7 but higher than that of a strong acid of the same concentration.
In conclusion, the pH of a 0.10 M benzoic acid solution is 2.59, confirming the weak acidic nature of benzoic acid.
