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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.