Question

What will be the titration curve and pH for titrating 0.2 M acetic acid with 25 ml 0.2 M of NaOH.

Answer 1

Question #64116, Chemistry / Other

What will be the titration curve and pH for titrating 0.2 M acetic acid with 25 ml 0.2 M of NaOH.

Answer:

For the beginning you have only CH₃COOH in the solution. Initial pH is a pH of the weak monoprotic acid solution:

$$\mathrm{CH_3COOH} = \mathrm{CH_3COO^-} + \mathrm{H^+}$$ $$K_a = \frac{[H^+]^2}{c_0 - [H^+]}$$ $$[H^+]^2 + [H^+]K_a - K_ac_0 = 0$$ $$c_0 = 0.2 \, \text{M}$$ $$K_a = 1.8 \times 10^{-5}$$ $$[H^+] = 1.89 \times 10^{-3}$$ $$pH = -\log[H^+] = -\log 1.89 \times 10^{-3} = 2.72$$

Equivalence point at 25 mL. All acid converted to salt:

$$\mathrm{CH_3COOH} + \mathrm{NaOH} = \mathrm{CH_3COONa} + \mathrm{H_2O}$$

This salt undergoes hydrolysis:

$$\mathrm{CH_3COO^-} + \mathrm{H_2O} = \mathrm{CH_3COOH} + \mathrm{OH^-}$$ $$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$ $$K_a = \frac{[OH^-]^2}{c_0 - [OH^-]}$$ $$[OH^-]^2 + [OH^-]K_b - K_b c_0 = 0$$ $$K_b = 5.6 \times 10^{-10}$$

Due to dilution:

$$c_0 = 0.1 \, \text{M}$$ $$[OH^-] = 7.48 \times 10^{-6}$$ $$pOH = -\log[OH^-] = -\log 7.48 \times 10^{-6} = 5.13$$ $$pH = 14 - pOH = 14 - 5.13 = 8.87$$

Before equivalence point pH will rise slowly, at equivalence point – sharp increase, after – again slow growth:

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