A sample of 50.0 mL of 0.10 M NH₃ ( K _b = 1.8×10^{- 5}) is titrated with 0.10 M HCl. Calculate the pH at the equivalence point.

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Quizplus · Accepted Answer

At the equivalence point in a titration between a weak base and a strong acid, the solution consists of the conjugate acid of the weak base. The pH is determined by the hydrolysis of this conjugate acid. The $K_b$ for the base is given, but we need the $K_a$ for its conjugate acid to calculate the pH. We use the relationship $K_w = K_a \times K_b$ where $K_w = 1.0 \times 10^{-14}$ at 25°C. Given $K_b = 1.8 \times 10^{-5}$, we find $K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$. Using the formula for pH of a solution of a weak acid, $pH = \frac{1}{2}(pK_a - \log[C_{acid}])$, and assuming the concentration of the conjugate acid is the same as the initial concentration of the base (0.10 M) since we are at the equivalence point, we find $pH = \frac{1}{2}(9.25 - \log[0.10]) = 5.28$.

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