Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100$M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\( with 0.100 \)\mathrm{M}$ \(\mathrm{HCl} .\)

The chemical reaction between NH3 and HCl can be written as: \(NH_{3}(aq) + HCl(aq) \rightarrow NH_{4}^{+}(aq) + Cl^{-}(aq)\) Initial conditions are as follows: Volume of \(NH_{3}\) solution: 25.0 mL Molarity of \(NH_{3}\) solution: 0.100 M \( K_{b} (NH_{3}) = 1.8 \times 10^{-5} \) Volume of \(HCl\) solution: To be determined Molarity of \(HCl\) solution: 0.100 M

Calculate the initial moles of NH3: moles of \(NH_{3}\) = volume × molarity moles of \(NH_{3}\) = (25.0 mL) × (0.100 mol/L) moles of \(NH_{3}\) = 0.00250 mol moles of \(HCl\) = 0 (Initially)

As the titration progresses, HCl is added to the solution and reacts with the NH3. At each stage of the titration, we will calculate the moles of NH3 and HCl in different scenarios. Scenario 1: Before adding any HCl moles of \(NH_{3}\) = 0.00250 mol moles of \(HCl\) = 0 mol Scenario 2: Halfway point (12.5 mL of HCl added) moles of \(HCl\) added = (12.5 mL) × (0.100 mol/L)= 0.00125 mol moles of \(NH_{3}\) remaining = 0.00250 mol - 0.00125 mol = 0.00125 mol Scenario 3: Equivalence point (25.0 mL of HCl added) moles of \(HCl\) added = (25.0 mL) × (0.100 mol/L) = 0.00250 mol moles of \(NH_{3}\) remaining = 0.00250 mol - 0.00250 mol = 0 mol moles of \(NH_{4}^{+}\) formed = 0.00250 mol Scenario 4: Past the equivalence point (30.0 mL of HCl added) moles of \(HCl\) added = (30.0 mL) × (0.100 mol/L) = 0.00300 mol moles of \(NH_{3}\) remaining = 0 moles of \(NH_{4}^{+}\) formed = 0.00250 mol excess moles of \(HCl\) = 0.00300 mol - 0.00250 mol = 0.00050 mol

In each scenario, we will use the moles of NH3, HCl, and NH4+ to find the pH of the solution. Scenario 1: Before adding any HCl Since HCl is not added yet, the pH of the solution depends on the basicity of NH3. Use \(K_{b}\) to calculate the concentration of \(OH^-\) formed, and then find the pH using the \(pOH\). Scenario 2: Halfway point (12.5 mL of HCl added) Use the Henderson-Hasselbalch equation to calculate the pH: \(pH = pK_{a} + log\frac{[base]}{[acid]}\) Here, \(pK_{a}\) is the negative logarithm of the acid dissociation constant for \(NH_{4}^{+}\), and the ratio will be the ratio of moles of NH3 to NH4+. Scenario 3: Equivalence point (25.0 mL of HCl added) Since all NH3 has reacted, we have only NH4+ in the solution. Calculate the concentration of \(NH_{4}^{+}\), use its dissociation constant to find the \(H^{+}\) concentration, and finally, calculate the pH. Scenario 4: Past the equivalence point (30.0 mL of HCl added) Calculate the excess moles of HCl, and find the \(H^{+}\) concentration in the solution. Then, calculate the pH. By following the steps above, you should be able to find the pH at each stage of the titration of NH3 with HCl.