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MaRussiya [10]
3 years ago
8

The following questions A24 - A26 relate to 100 ml of 0.0150 M solution of benzoic acid

Chemistry
1 answer:
pogonyaev3 years ago
8 0

Answer:

PH of the weak acid: approximately 2.513.

PH of the buffer solution: approximately 4.495.

Explanation:

The Ka value of benzoic acid is much smaller than 1. Benzoic acid will dissociate but only partially when dissolved in water. Construct a RICE table for this process. Let the equilibrium of \rm H^{+} be x\; \rm mol\cdot L^{-1}. Note that x \ge 0.

\displaystyle \begin{array}{c|ccccc}\textbf{R}&\mathrm{C_6H_5COOH} & \rightleftharpoons & \mathrm{C_6H_5COO^{-}} & + &\mathrm{H^{+}}\\\textbf{I} & 0.015 & & 0 & & 0\\\textbf{C} & -x & & +x & & +x\\\textbf{E} & 0.015-x & & x & & x\end{array}

\displaystyle \frac{[\mathrm{C_6H_5COO^{-}}]\cdot [\mathrm{H^{+}}]}{[\mathrm{C_6H_5COOH}]} = \mathrm{pK}_{a}.

\displaystyle \frac{x^{2}}{0.015 - x} = 6.4\times 10^{-5}.

Solve this quadratic equation for x:

x^{2}+ 6.4\times 10^{-5}\;x - 6.4\times 10^{-5}\times 0.150 = 0.

\displaystyle x = \frac{-6.4\times 10^{-5} \pm \sqrt{(6.4\times 10^{-5})^{2} - 4\times (- 6.4\times 10^{-5}\times 0.150)}}{2}.

Take only the non-negative root. x \approx 0.00306655.

\rm [H^{+}] = 0.00306655\; mol\cdot L^{-1}.

\displaystyle \mathrm{pH} = -\log_{10}{[\mathrm{H^{+}}]} = 2.513.

Each benzoic acid contains only one carboxyl group \mathrm{-COOH}. Benzoic acid is thus a monoprotic acid. Each mole of the acid will react with only one mole of \rm NaOH. The 100 mL solution initially contains 1.50\times 10^{-3} moles of benzoic acid. The 1\times 10^{-3} moles of \rm NaOH will neutralize only part of the acid. The solution will eventually contain 1\times 10^{-3} moles of \mathrm{C_6H_5COO^{-}} (from the salt \mathrm{C_6H_5COONa}) and 0.50\times 10^{-3} moles of \mathrm{C_6H_3COOH}.

Both the acid \mathrm{C_6H_5COOH} and the conjugate base of the acid \mathrm{C_6H_5COO^{-}} exist in large amounts in the solution. Apply the Henderson-Hasselbalch equation for weak acid buffers to find the pH of this buffer solution.

\mathrm{pK}_{a} = -\log_{10}{\mathrm{K}_{a}} \approx 4.19382 for benzoic acid.

\begin{aligned}\mathrm{pH} &= \mathrm{pK}_{a} + \log{\frac{{[\text{Conjugate Base}]}}{[\text{Weak Acid}]}} \\ &= \mathrm{pK}_{a} + \log{\frac{{[\mathrm{C_6H_5COO^{-}}]}}{[\mathrm{C_6H_5COOH}]}}\\ &= 4.19382 + \log{\frac{0.01}{0.005}}\\ &\approx 4.495 \end{aligned}.

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