Question
A 20.0mL aliquot of 0.100 M HCl is titrated with 0.100M NaOH solution. What is the pH of the resulting solution after 15.0mL of NaOH has been added?
Answers
GPT-4o mini
To determine the pH of the solution after adding 15.0 mL of 0.100 M NaOH to 20.0 mL of 0.100 M HCl, we first need to figure out how many moles of HCl and NaOH are present.
1. Calculate moles of HCl:
\[
\text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.0200 \, \text{L} \times 0.100 \, \text{M} = 0.00200 \, \text{mol}
\]
2. Calculate moles of NaOH added:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.0150 \, \text{L} \times 0.100 \, \text{M} = 0.00150 \, \text{mol}
\]
3. Determine the reaction and the remaining moles after reaction:
The neutralization reaction is:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
After 15.0 mL of NaOH is added, the moles of HCl and NaOH will react in a 1:1 molar ratio:
- Initially, we have 0.00200 mol of HCl and add 0.00150 mol of NaOH.
- Moles of HCl remaining after reaction:
\[
0.00200 \, \text{mol HCl} - 0.00150 \, \text{mol NaOH} = 0.00050 \, \text{mol HCl}
\]
- Moles of NaOH remaining:
Since all of the NaOH reacts, the moles of NaOH left is:
\[
0.00150 \, \text{mol NaOH} - 0.00150 \, \text{mol NaOH} = 0 \, \text{mol}
\]
Since there is still HCl left in the solution, the solution is acidic. We will calculate the concentration of the remaining HCl.
4. Calculate the total volume of the solution:
Total volume after addition of NaOH:
\[
20.0 \, \text{mL HCl} + 15.0 \, \text{mL NaOH} = 35.0 \, \text{mL} = 0.0350 \, \text{L}
\]
5. Calculate concentration of the remaining HCl:
\[
\text{Concentration of HCl} = \frac{0.00050 \, \text{mol}}{0.0350 \, \text{L}} \approx 0.01429 \, \text{M}
\]
6. Calculate the pH of the solution:
For a strong acid like HCl, the pH can be calculated directly from the concentration of HCl:
\[
\text{pH} = -\log[H^+] = -\log(0.01429) \approx 1.84
\]
Thus, the pH of the resulting solution after adding 15.0 mL of NaOH to the HCl solution is approximately **1.84**.
1. Calculate moles of HCl:
\[
\text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.0200 \, \text{L} \times 0.100 \, \text{M} = 0.00200 \, \text{mol}
\]
2. Calculate moles of NaOH added:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.0150 \, \text{L} \times 0.100 \, \text{M} = 0.00150 \, \text{mol}
\]
3. Determine the reaction and the remaining moles after reaction:
The neutralization reaction is:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
After 15.0 mL of NaOH is added, the moles of HCl and NaOH will react in a 1:1 molar ratio:
- Initially, we have 0.00200 mol of HCl and add 0.00150 mol of NaOH.
- Moles of HCl remaining after reaction:
\[
0.00200 \, \text{mol HCl} - 0.00150 \, \text{mol NaOH} = 0.00050 \, \text{mol HCl}
\]
- Moles of NaOH remaining:
Since all of the NaOH reacts, the moles of NaOH left is:
\[
0.00150 \, \text{mol NaOH} - 0.00150 \, \text{mol NaOH} = 0 \, \text{mol}
\]
Since there is still HCl left in the solution, the solution is acidic. We will calculate the concentration of the remaining HCl.
4. Calculate the total volume of the solution:
Total volume after addition of NaOH:
\[
20.0 \, \text{mL HCl} + 15.0 \, \text{mL NaOH} = 35.0 \, \text{mL} = 0.0350 \, \text{L}
\]
5. Calculate concentration of the remaining HCl:
\[
\text{Concentration of HCl} = \frac{0.00050 \, \text{mol}}{0.0350 \, \text{L}} \approx 0.01429 \, \text{M}
\]
6. Calculate the pH of the solution:
For a strong acid like HCl, the pH can be calculated directly from the concentration of HCl:
\[
\text{pH} = -\log[H^+] = -\log(0.01429) \approx 1.84
\]
Thus, the pH of the resulting solution after adding 15.0 mL of NaOH to the HCl solution is approximately **1.84**.