To solve the problem step by step, we will follow these calculations:
### Step 1: Calculate the number of millimoles of NH₃
Given:
- Volume of NH₃ solution = 20 mL
- Molarity of NH₃ = 0.1 M
Using the formula:
\[
\text{Number of millimoles} = \text{Molarity} \times \text{Volume (mL)}
\]
\[
\text{Number of millimoles of NH₃} = 0.1 \, \text{mol/L} \times 20 \, \text{mL} = 2 \, \text{mmol}
\]
### Step 2: Calculate the number of millimoles of HCl
Given:
- Molarity of HCl = 0.025 M
- Let the volume of HCl be \( V \) mL.
Using the formula:
\[
\text{Number of millimoles of HCl} = 0.025 \, \text{mol/L} \times V \, \text{mL}
\]
### Step 3: Determine the volume of HCl required for neutralization
At the equivalence point, the number of millimoles of NH₃ will equal the number of millimoles of HCl:
\[
2 \, \text{mmol} = 0.025 \, \text{mol/L} \times V
\]
Solving for \( V \):
\[
V = \frac{2 \, \text{mmol}}{0.025 \, \text{mol/L}} = 80 \, \text{mL}
\]
### Step 4: Write the neutralization reaction
The reaction between NH₃ and HCl is:
\[
\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}
\]
At the equivalence point, all NH₃ is converted to NH₄Cl.
### Step 5: Calculate the concentration of NH₄Cl
Total volume of the solution after mixing NH₃ and HCl:
\[
\text{Total volume} = 20 \, \text{mL} + 80 \, \text{mL} = 100 \, \text{mL}
\]
The concentration of NH₄Cl:
\[
\text{Concentration of NH₄Cl} = \frac{2 \, \text{mmol}}{100 \, \text{mL}} = 0.02 \, \text{mol/L} = 0.1 \, \text{M}
\]
### Step 6: Calculate pH at the equivalence point
At the equivalence point, we have a solution of NH₄Cl, which is a weak acid. We need to find the pH using the \( K_b \) of NH₃.
Given:
- \( K_b \) of NH₃ = \( 2 \times 10^{-6} \)
First, calculate \( K_a \) for NH₄⁺:
\[
K_w = K_a \times K_b \Rightarrow K_a = \frac{K_w}{K_b}
\]
Where \( K_w = 1 \times 10^{-14} \) at 25°C.
\[
K_a = \frac{1 \times 10^{-14}}{2 \times 10^{-6}} = 5 \times 10^{-9}
\]
Now, use the formula for pH:
\[
\text{pH} = 7 + \frac{1}{2} pK_a - \frac{1}{2} \log C
\]
Where \( C = 0.1 \, \text{M} \) (concentration of NH₄Cl).
Calculate \( pK_a \):
\[
pK_a = -\log(5 \times 10^{-9}) \approx 8.3
\]
Now substitute into the pH formula:
\[
\text{pH} = 7 + \frac{1}{2}(8.3) - \frac{1}{2}(\log(0.1))
\]
\[
\text{pH} = 7 + 4.15 + 0.5 = 11.65
\]
### Final Answer:
The pH of the reaction mixture at the equivalence point is approximately **11.65**.
