To find the pH of the solution after mixing NaOH and HCl, we need to first determine the number of moles of each.
1. Calculate moles of NaOH:
- Volume of NaOH = 49.25 mL = 0.04925 L
- Concentration of NaOH = 0.080 M
\[ \text{Moles of NaOH} = 0.080 , \text{mol/L} \times 0.04925 , \text{L} = 0.00394 , \text{mol} \]
2. Calculate moles of HCl:
- Volume of HCl = 260.9 mL = 0.2609 L
- Concentration of HCl = 0.060 M
\[ \text{Moles of HCl} = 0.060 , \text{mol/L} \times 0.2609 , \text{L} = 0.01565 , \text{mol} \]
3. Determine the limiting reactant: The reaction between NaOH and HCl is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Moles of HCl = 0.01565 mol
- Moles of NaOH = 0.00394 mol
Since there are fewer moles of NaOH, it is the limiting reactant.
4. Calculate remaining moles after reaction:
- Moles of HCl remaining = \(0.01565 - 0.00394 = 0.01171 , \text{mol}\)
- Moles of NaOH remaining = 0 (all NaOH is used up)
5. Calculate the concentration of HCl remaining in the total volume:
- Total volume = 49.25 mL + 260.9 mL = 310.15 mL = 0.31015 L
\[ \text{Concentration of remaining HCl} = \frac{0.01171 , \text{mol}}{0.31015 , \text{L}} \approx 0.0377 , \text{M} \]
6. Calculate the pH: Using the formula for pH:
\[ \text{pH} = -\log[H^+] = -\log[0.0377] \approx 1.43 \]
The final answer, to two decimal places, is:
pH = 1.43