We are dealing with a titration problem involving ammonia ($\text{NH}_3$) as the weak base and hydrochloric acid (HCl) as the strong acid. Let's break the problem down into two parts:

Part (a) - Volume of HCl to reach the equivalence point:

The equivalence point in a titration is reached when the moles of acid equal the moles of base. First, calculate the moles of ammonia in the initial solution.

1. Moles of NH₃: $\text{Moles of NH₃} = M_{\text{NH₃}} \times V_{\text{NH₃}} = 0.044 \, M \times 0.075 \, L = 0.0033 \, \text{mol}$

   The moles of HCl needed to reach the equivalence point will be the same because the reaction between NH₃ and HCl proceeds in a 1:1 molar ratio.

2. Volume of HCl:

   The moles of HCl can be calculated using the molarity and volume of HCl: $\text{Volume of HCl} = \frac{\text{Moles of HCl}}{M_{\text{HCl}}} = \frac{0.0033 \, \text{mol}}{0.087 \, M} = 0.03793 \, L = 37.93 \, \text{mL}$

Thus, the volume of $0.087 \, M$ HCl required to reach the equivalence point is 37.93 mL.

Part (b) - pH at the equivalence point:

At the equivalence point, all the ammonia ($\text{NH}_3$) has reacted with HCl to form ammonium ions ($\text{NH}_4^+$). The pH is determined by the hydrolysis of $\text{NH}_4^+$, which acts as a weak acid.

1. Concentration of $\text{NH}_4^+$ after the equivalence point:

   The total volume after adding the acid is: $V_{\text{total}} = V_{\text{NH}_3} + V_{\text{HCl}} = 75 \, \text{mL} + 37.93 \, \text{mL} = 112.93 \, \text{mL} = 0.11293 \, \text{L}$

   The concentration of $\text{NH}_4^+$ is then: $[\text{NH}_4^+] = \frac{0.0033 \, \text{mol}}{0.11293 \, \text{L}} = 0.0292 \, M$

2. Hydrolysis of $\text{NH}_4^+$:

   The reaction of $\text{NH}_4^+$ with water is: $\text{NH}_4^+ + H_2O \rightleftharpoons \text{NH}_3 + H_3O^+$

   The equilibrium constant for this reaction is related to the base dissociation constant $K_b$ of ammonia by: $K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

3. Calculation of $\text{H}^+$ concentration:

   Set up the expression for the ionization of $\text{NH}_4^+$ to find the concentration of $\text{H}^+$: [ K_a = \frac{[\text{H}^+][\text{NH}_3]}{[\text{NH}_4^+]} ] Assume $[\text{H}^+] = x$ and $[\text{NH}_4^+] = 0.0292 \, M$, we get: $5.56 \times 10^{-10} = \frac{x^2}{0.0292}$ Solving for $x$, we find: $x^2 = (5.56 \times 10^{-10}) \times (0.0292) = 1.622 \times 10^{-11}$ $x = \sqrt{1.622 \times 10^{-11}} = 1.273 \times 10^{-6} \, M$

   Therefore, $[\text{H}^+] = 1.273 \times 10^{-6} \, M$.

4. pH Calculation: $\text{pH} = -\log[\text{H}^+] = -\log(1.273 \times 10^{-6}) = 5.90$

Thus, the pH at the equivalence point is 5.90.

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Do you have any questions or want further details on any of these calculations?

Here are 5 related questions for you to think about:

1. How would the pH change if a stronger base than ammonia were used?
2. How does the volume of HCl required change with increasing ammonia concentration?
3. What happens to the pH of the solution if an excess of HCl is added beyond the equivalence point?
4. How would the titration curve for this reaction look?
5. How does the value of $K_b$ influence the pH at the equivalence point?

Tip: In weak base-strong acid titrations, the equivalence point pH is always acidic due to the hydrolysis of the weak base's conjugate acid.