Question

Calculate the pH at equivalence point for the titration of 0.220M methylamine with 0.220M of HCl. The Kb of methylamine is 5.0x10^-4

Answer 1

Answer on Question #48074 - Chemistry - Inorganic Chemistry

Question

Calculate the pH at equivalence point for the titration of 0.220M methylamine with 0.220M of HCl. The Kb of methylamine is $5.0 \times 10^{-4}$

Answer:

The reaction of titration of methylamine with HCl:

$$\mathrm{CH_3NH_2} + \mathrm{HCl} \rightleftharpoons \mathrm{CH_3NH_3^+} + \mathrm{Cl^-}$$

The equivalence point is a point where the moles of hydrochloric acid added is equal to initial moles of methylamine in solution.

$$K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{5.0 \times 10^{-4}} = 2.0 \cdot 10^{-11}$$

After equivalence point (x – number of $\mathrm{CH_3NH_3^+}$ moles reacted):

$$\mathrm{CH_3NH_3^+} + \mathrm{H_2O} \rightarrow \mathrm{H_3O^+} + \mathrm{CH_3NH_2}$$

at end $0.220 - x$ x x

$$K_a = \frac{x \cdot x}{0.220 - x} = 2.0 \cdot 10^{-11}$$ $$x^2 - 4.40 \cdot 10^{-12} + 2.0 \cdot 10^{-11} x = 0$$ $$x = 2.1 \cdot 10^{-6}$$

So, concentration of $\mathrm{H_3O^+}$ is $2.1 \cdot 10^{-6}$ M. Therefore, pH at equivalence point is:

$$\mathrm{pH} = -\log[\mathrm{H_3O^+}] = -\log(2.1 \cdot 10^{-6}) = 5.68$$

Answer: $\mathrm{pH} = 5.68$

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