Question

Calculate the pH of the resulting solution if 16.0 mL. of 0.160 M HCl(aq) is added to 21.0 mL of 0.160 M NaOH(aq

Answer 1

To calculate the pH of the resulting solution when 16.0 mL of 0.160 M HCl is mixed with 21.0 mL of 0.160 M NaOH, we can follow these steps:

1. **Calculate the moles of HCl and NaOH:**

- Moles of HCl:
\[
\text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.016 \, \text{L} \times 0.160 \, \text{M} = 0.00256 \, \text{moles}
\]

- Moles of NaOH:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.021 \, \text{L} \times 0.160 \, \text{M} = 0.00336 \, \text{moles}
\]

2. **Determine which reactant is in excess:**

- The balanced reaction for HCl and NaOH is:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]

Since both HCl and NaOH react in a 1:1 molar ratio, we can compare their moles:

- \( \text{Moles of NaOH} = 0.00336 \)
- \( \text{Moles of HCl} = 0.00256 \)

Since \( 0.00336 > 0.00256 \), NaOH is in excess.

3. **Calculate the moles of excess NaOH:**
\[
\text{Excess NaOH} = \text{Moles of NaOH} - \text{Moles of HCl} = 0.00336 - 0.00256 = 0.00080 \, \text{moles of NaOH}
\]

4. **Calculate the total volume of the solution:**
\[
\text{Total Volume} = 16.0 \, \text{mL} + 21.0 \, \text{mL} = 37.0 \, \text{mL} = 0.037 \, \text{L}
\]

5. **Calculate the concentration of excess NaOH:**
\[
\text{Concentration of excess NaOH} = \frac{\text{Moles of excess NaOH}}{\text{Total Volume (L)}} = \frac{0.00080 \, \text{moles}}{0.037 \, \text{L}} \approx 0.0216 \, \text{M}
\]

6. **Calculate the pOH and then the pH:**
\[
\text{pOH} = -\log[\text{OH}^-] = -\log(0.0216) \approx 1.666
\]
\[
\text{pH} = 14 - \text{pOH} \approx 14 - 1.666 \approx 12.334
\]

Thus, the pH of the resulting solution is approximately **12.33**.