Answer:
To calculate the pH at the equivalence point of the titration, we need to find the concentration of the resulting solution after the reaction between NH3 and HCl. At the equivalence point, the moles of NH3 and HCl are equal.
Given:
- Volume of NH3 (VB) = 10.0 mL
- Concentration of NH3 (CB) = 0.18 M
- Volume of HCl (VA) = 10.0 mL
- Concentration of HCl (CA) = 0.18 M
- \( K_a \) of NH4^+ = \( 5.8 \times 10^{-10} \)
1. Calculate the moles of NH3 and HCl:
\[ \text{Moles of NH}_3 = \text{CB} \times \text{VB} \]
\[ = 0.18 \, \text{M} \times 0.010 \, \text{L} \]
\[ = 0.0018 \, \text{moles} \]
Since NH3 and HCl react in a 1:1 ratio, the moles of HCl are also 0.0018 moles.
2. Determine the volume of the solution after mixing NH3 and HCl:
Total volume = VB + VA = 10.0 mL + 10.0 mL = 20.0 mL = 0.020 L
3. Calculate the concentration of the resulting solution:
Total moles = moles of NH3 + moles of HCl = 0.0018 moles + 0.0018 moles = 0.0036 moles
\[ \text{Concentration} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.0036 \, \text{moles}}{0.020 \, \text{L}} \]
\[ = 0.18 \, \text{M} \]
4. Since NH3 is a weak base, it reacts with water to produce NH4^+ and OH^- ions. We need to find the concentration of OH^- ions at the equivalence point:
\[ K_b = \frac{K_w}{K_a} \]
\[ K_b = \frac{1.0 \times 10^{-14}}{5.8 \times 10^{-10}} \]
\[ K_b \approx 1.72 \times 10^{-5} \]
At the equivalence point, \( [\text{OH}^-] = \sqrt{K_b \times \text{CB}} \)
\[ [\text{OH}^-] = \sqrt{1.72 \times 10^{-5} \times 0.18} \]
\[ [\text{OH}^-] \approx 3.71 \times 10^{-3} \, \text{M} \]
5. Since \( [\text{H}^+] \times [\text{OH}^-] = K_w = 1.0 \times 10^{-14} \):
\[ [\text{H}^+] = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]} \]
\[ [\text{H}^+] = \frac{1.0 \times 10^{-14}}{3.71 \times 10^{-3}} \]
\[ [\text{H}^+] \approx 2.69 \times 10^{-12} \, \text{M} \]
6. Calculate the pH:
\[ \text{pH} = -\log{[\text{H}^+]} \]
\[ \text{pH} = -\log{(2.69 \times 10^{-12})} \]
\[ \text{pH} \approx 11.57 \]
So, the pH of the solution at the equivalence point is approximately 11.57.
