Question

Calculate the hydronium ion concentration and pH at the equivalence point in a titration of $50.0mL$ of $0.40M{NH}_3$ with $0.40M$ HCl. $K_b=2X{10}^{-5}$

Answer 1

${NH}_3+HCl⟶{NH}_{4}Cl$

${NH}_{4}Cl+H_{2}O\to H_3{O}^{+}+N^{(-)}+{NH}_3$

$K_n=\frac{K_w}{K_b}$

$\left[H^{+}\right]=\sqrt{\frac{K_{w}C}{K_b}}$

$C=\frac{50\times 40\times {10}^{-2}}{100}0.2M$

$\left[H^{+}\right]=\sqrt{\frac{{10}^{-14}\times 0.2}{2\times {10}^{-5}}}=\sqrt{{10}^{-9}\times 0.1}={10}^{-5}$

$pH=5$