A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

We are given the concentration of ammonia, \(0.400\,M\). In order to find the moles of ammonia, we need to know the volume of the solution. We're told the total volume at the equivalence point was 1.50 times the original volume. Since the volume of ammonia and hydrochloric acid combined is 1.50 times the original volume, the volume of the ammonia must be half of the total volume. Let's denote the volume of ammonia as \(V_{NH_3}\), then we have: \(2 \times V_{NH_3} = 1.50 V_{NH_3}\) Solving for \(V_{NH_3}\), we get: \(V_{NH_3} = 0.75 V_{NH_3}\) Now, we can find the moles of ammonia using the provided concentration: moles \(NH_3 = 0.400\,M \times 0.75 V_{NH_3}\)

At the equivalence point, the moles of ammonia and hydrochloric acid are equal. Therefore, the moles of hydrochloric acid are the same as the moles of ammonia. So: moles \(HCl = 0.400\,M \times 0.75 V_{NH_3}\)

Since ammonia is a weak base and hydrochloric acid is a strong acid, the product of the reaction is the conjugate acid of ammonia, which is ammonium chloride (\(NH_4Cl\)). \(NH_3 + HCl \rightarrow NH_4Cl\) At the equivalence point, all the ammonia has reacted with the hydrochloric acid to form ammonium ions (\(NH_4^+\)). The equilibrium reaction for the ammonium ion is: \(NH_4^+ \rightleftharpoons NH_3 + H^+\) To calculate the pH at the equivalence point, we need to find the equilibrium concentration of \(H^+\) ions. We can do this using the equilibrium constant expression for the reaction: \(K_a = \frac{[NH_3][H^+]}{[NH_4^+]}\) We are given the initial concentration of ammonia (0.400 M), and we know that it has reacted with an equal amount of hydrochloric acid. Therefore, the initial concentration of ammonium ion is also 0.400 M. Since the reaction is a 1:1 ratio, the final concentration of \(NH_3\) and \(H^+\) at equilibrium will be the same. Thus, we can simplify the \(K_a\) expression as: \(K_a = \frac{x^2}{0.400\,M-x}\) The \(K_a\) value for ammonia is approximately \(5.56 \times 10^{-10}\). By substituting this value into the equation above and solving for x, we can find the concentration of \(H^+\) ions at equilibrium.

\(5.56 \times 10^{-10} = \frac{x^2}{0.400\,M-x}\) Solve for x (concentration of \(H^+\)) using either a quadratic equation or an approximation method. For simplicity, we can approximate that x is much smaller than 0.400 M so the equation becomes: \(5.56 \times 10^{-10} \approx \frac{x^2}{0.400\,M}\) Solve for x: \(x \approx 1.18 \times 10^{-5}\,M\)

Now that we have the equilibrium concentration of \(H^+\) ions, we can use the pH formula to find the pH at the equivalence point: pH = -log\([H^+]\) pH = -log\((1.18 \times 10^{-5}\,M)\) pH ≈ 4.93 At the equivalence point, the pH is approximately 4.93.