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Chapter 15: Problem 43

The \(K_{\mathrm{a}}\) for benzoic acid is \(6.5 \times 10^{-5} .\) Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) benzoic acid solution.

Short Answer

Expert verified
The pH of a 0.10 M benzoic acid solution is approximately 2.59.

Step by step solution

01

Write the Ionization Equation of Benzoic Acid

The ionization or dissociation of benzoic acid in water can be represented as follows: \(C_6H_5COOH ⇌ C_6H_5COO^- + H^+\)
02

Apply the Definition of \(K_a\)

The acid dissociation constant, \(K_a\), is given by the formula: \(K_a = \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}\). At equilibrium, the concentrations of \(C_6H_5COO^- \) and \(H^+\) are equal and can be written as 'x'. Thus, the concentration of \(C_6H_5COOH\) is \((0.10 - x)\). Given that \(K_a = 6.5 \times 10^{-5} \), the equation becomes: \(6.5 \times 10^{-5} = \frac{(x)(x)}{0.10 - x}\).
03

Solve for \(x\)

Assume that \(x\) is much smaller than 0.10 M, so that \(0.10 - x ≈ 0.10\) simplifying the equation to: \(6.5 \times 10^{-5} = \frac{x^2}{0.10}\). Solving for \(x\) which represents the [H+], we get \(x = \sqrt{(0.10)(6.5 \times 10^{-5})} = 2.55 \times 10^{-3} \mathrm{M}\)
04

Calculate pH

Finally, we apply the definition of pH, which is \(pH=\-log[H^{+}]\). Substituting \(2.55 \times 10^{-3} \mathrm{M}\) for \(H^{+}\), the pH is \( pH = -\log(2.55 \times 10^{-3}) ≈ 2.59\).

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Most popular questions from this chapter

Both the amide ion \(\left(\mathrm{NH}_{2}^{-}\right)\) and the nitride ion \(\left(\mathrm{N}^{3-}\right)\) are stronger bases than the hydroxide ion and hence do not exist in aqueous solutions. (a) Write equations showing the reactions of these ions with water, and identify the Brønsted acid and base in each case. (b) Which of the two is the stronger base?

What are the concentrations of \(\mathrm{HSO}_{4}^{-}, \mathrm{SO}_{4}^{2-},\) and \(\mathrm{H}^{+}\) in a \(0.20 \mathrm{M} \mathrm{KHSO}_{4}\) solution?

Calculate the hydrogen ion concentration in \(\mathrm{mol} / \mathrm{L}\) for each of the following solutions: (a) a solution whose \(\mathrm{pH}\) is \(5.20,\) (b) a solution whose \(\mathrm{pH}\) is 16.00 , (c) a solution whose hydroxide concentration is \(3.7 \times 10^{-9} M\).

Arrange the oxides in each of the following groups in order of increasing basicity: (a) \(\mathrm{K}_{2} \mathrm{O}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{BaO}\) (b) \(\mathrm{CrO}_{3}, \mathrm{CrO}, \mathrm{Cr}_{2} \mathrm{O}_{3}\)

Henry's law constant for \(\mathrm{CO}_{2}\) at \(38^{\circ} \mathrm{C}\) is \(2.28 \times 10^{-3}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{atm} .\) Calculate the \(\mathrm{pH}\) of a solution of \(\mathrm{CO}_{2}\) at \(38^{\circ} \mathrm{C}\) in equilibrium with the gas at a partial pressure of 3.20 atm.
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