Question

calculate the pH of 1.0 M solution of acetic acid. To what volume one litre of this solution be diluted so that the pH of the solution that is formed will be twice of original volume.[Ka = 1.8 ×10^-5]

Answer 1

Question#46462 – Chemistry – Inorganic Chemistry

Question:

Calculate the pH of 1.0 M solution of acetic acid. To what volume one liter of this solution be diluted so that the pH of the solution that is formed will be twice of original volume. [Kₐ = 1.8 × 10⁻⁵]

Answer:

Acetic acid CH₃COOH is a weak acid and it dissociated in water solution to some extent according to equation:

$$\mathrm{CH_3COOH} \leftrightarrow \mathrm{CH_3COO^-} + \mathrm{H^+}$$

Since the process is reversible, the constant of equilibrium of this process $K_{a} = \frac{\left[H^{+}\right]\left[CH_{3}COO^{-}\right]}{\left[CH_{3}COOH\right]}$

The degree of dissociation for CH₃COOH is not great, than we can neglect the amount of CH₃COOH that was ionized comparing with the initial concentration. And, according to the reaction equation, the amount of H⁺ and CH₃COO⁻ formed are the same. Using this consideration:

$$K_{a} = \frac{\left[H^{+}\right]^{2}}{\left[CH_{3}COOH\right]_{0}} \Rightarrow \sqrt{\left[H^{+}\right]} = K_{a} \times \left[CH_{3}COOH\right]_{0}$$

We have, that $\left[\mathrm{CH_3COOH}\right]_0 = 1.0\,\mathrm{M}$ and $K_{a} = 1.8 \cdot 10^{-5}$. Therefore,

$$\left[H^{+}\right] = \sqrt{K_{a} \times \left[CH_{3}COOH\right]_{0}} = \sqrt{1.8 \cdot 10^{-5} \times 1.0} = 0.0042\,\mathrm{M}$$

pH function is a negative logarithm from $\left[\mathrm{H^{+}}\right]$:

$$pH = -\log \left[H^{+}\right] = -\log (0.0042) = 2.38$$

pH after the dilution has to be twice of original volume pH = 2.38 × 2 = 4.76. The corresponding H⁺ concentration is:

$$pH = -\log \left[H^{+}\right] \Rightarrow \left[H^{+}\right] = 10^{-pH} = 10^{-4.76} = 1.74 \cdot 10^{-5}\,\mathrm{M}$$

The corresponding initial concentration of CH₃COOH, which produced this amount of H⁺:

$$\left[H^{+}\right] = \sqrt{K_{a} \times \left[CH_{3}COOH\right]_{0}} \Rightarrow \left[CH_{3}COOH\right]_{0} = \frac{\left[H^{+}\right]^{2}}{K_{a}} = \frac{\left(1.74 \cdot 10^{-5}\right)^{2}}{1.8 \cdot 10^{-5}} = 1.68 \cdot 10^{-5}\,\mathrm{M}$$

The amount of moles remains constant, only volume of the solution has changed. If the initial volume $V_{1} = 1$ L, hence:

$$C_{1}V_{1} = C_{2}V_{2} \Rightarrow V_{2} = \frac{C_{1}V_{1}}{C_{2}} = \frac{1 \times 1.0}{1.68 \cdot 10^{-5}} \approx 6 \cdot 10^{5}\,\mathrm{L}$$

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