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?

Ammonia (NH3) reacts with hydrochloric acid (HCl) to form ammonium chloride (NH4Cl). The balanced equation is: \( NH_3 + HCl \rightarrow NH_4Cl \)

We are given the concentration of ammonia (0.400 mol/L) and the original volume (which we will call V). To find the moles of ammonia, we need to multiply the concentration by the volume: Moles of NH3 = concentration × volume = 0.400 mol/L × V

At the equivalence point, the moles of HCl added are equal to the moles of NH3 in the original solution. From Step 2, the moles of NH3 = 0.400 mol/L × V. So, the moles of HCl added are: Moles of HCl = 0.400 mol/L × V

The total volume at the equivalence point is given as 1.50 times the original volume. Therefore, Total Volume = 1.50 × V

At the equivalence point, all moles of NH3 have reacted with HCl, and an equal number of moles of NH4+ ions have been produced. The concentration of NH4+ ions can be calculated using the total volume at the equivalence point: Concentration of NH4+ = moles of NH4+ / total volume Concentration of NH4+ = (0.400 mol/L × V) / (1.50 × V) We see that the volume V cancels out: Concentration of NH4+ = (0.400 mol/L) / 1.50

At the equivalence point, the solution is composed of NH4+ ions, which can hydrolyze water to produce H3O+ ions and NH3. We can write the equilibrium expression (Kb) for the reaction as follows: \( NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq) \) \(K_b = \frac{[NH_3][H_3O^+]}{[NH_4^+]} \) Given that the concentration of NH4+ ions at the equivalence point has already been determined in Step 5, we can use the relationship \(K_w = K_a \times K_b\) to find the Kb, where Kw is the ion product constant for water (1.0 × 10-14) and Ka is the acid ionization constant for NH4+. Rearrange the equation to find Kb: \(K_b = \frac{K_w}{K_a} \) Now, we can use the Kb expression to solve for the concentration of H3O+ ions: \([H_3O^+] = \sqrt{K_b \times [NH_4^+]} \) Finally, use the concentration of H3O+ ions to calculate the pH: pH = -log10([H3O+])

We have all the necessary values to calculate the pH at the equivalence point. Substitute the values into the equations from Step 6 and compute the final pH value. This will give us the pH of the solution at the equivalence point.