Question

Calculate the pH of the resulting solution if 16.0 ml of 0.160 M HCl(aq) is added to 26.0 ml. of 0.210 MNOH

Answer 1

To calculate the pH of the resulting solution after mixing \(16.0 \, \text{mL}\) of \(0.160 \, \text{M HCl}\) with \(26.0 \, \text{mL}\) of \(0.210 \, \text{M NOH}\), we need to follow these steps:

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

- For HCl:
\[
\text{moles of HCl} = \text{Volume in L} \times \text{Concentration} = 0.016 \, \text{L} \times 0.160 \, \text{mol/L} = 0.00256 \, \text{mol}
\]

- For NOH:
\[
\text{moles of NOH} = \text{Volume in L} \times \text{Concentration} = 0.026 \, \text{L} \times 0.210 \, \text{mol/L} = 0.00546 \, \text{mol}
\]

2. **Determine the reaction between HCl and NOH:**

HCl is a strong acid and NOH is a strong base. The neutralization reaction is:
\[
\text{HCl} + \text{NOH} \rightarrow \text{H}_2\text{O} + \text{NOCl}
\]

3. **Find the limiting reagent:**

Since we have \(0.00256 \, \text{mol HCl}\) and \(0.00546 \, \text{mol NOH}\), HCl is the limiting reagent.

4. **Calculate the moles of acid and base remaining:**

Moles of NOH after reaction:
\[
\text{Remaining NOH} = 0.00546 \, \text{mol} - 0.00256 \, \text{mol} = 0.00290 \, \text{mol}
\]

Moles of HCl after reaction:
\[
\text{Remaining HCl} = 0\, \text{mol} \, \text{(since it's the limiting reagent)}
\]

5. **Calculate the concentration of NOH remaining:**

Total volume of the resulting solution:
\[
\text{Total volume} = 16.0 \, \text{mL} + 26.0 \, \text{mL} = 42.0 \, \text{mL} = 0.042 \, \text{L}
\]

Concentration of remaining NOH:
\[
\text{Concentration of NOH} = \frac{0.00290 \, \text{mol}}{0.042 \, \text{L}} = 0.06905 \, \text{M}
\]

6. **Calculate the pOH of the solution:**

\[
\text{pOH} = -\log[\text{NOH}] = -\log[0.06905] \approx 1.16
\]

7. **Calculate the pH of the solution:**

Since \( \text{pH} + \text{pOH} = 14 \):
\[
\text{pH} = 14 - 1.16 \approx 12.84
\]

**Final Answer:**
The pH of the resulting solution is approximately **12.84**.