Sure, let's work through the problem of calculating the pH of the solution after 10 ml of 0.05M NaOH solution has been added to 100 ml of 0.1M HNO₃ solution.
First, let's determine the moles of HNO₃ and NaOH.
1. Calculate moles of HNO₃:
- Volume of HNO₃ = 100 ml = 0.1 L
- Molarity of HNO₃ = 0.1 M
- Moles of HNO₃ = volume (L) * molarity (M)
[tex]\[
\text{Moles of HNO}_3 = 0.1 \times 0.1 = 0.01 \text{ moles}
\][/tex]
2. Calculate moles of NaOH:
- Volume of NaOH = 10 ml = 0.01 L
- Molarity of NaOH = 0.05 M
- Moles of NaOH = volume (L) * molarity (M)
[tex]\[
\text{Moles of NaOH} = 0.01 \times 0.05 = 0.0005 \text{ moles}
\][/tex]
3. Determine the moles of HNO₃ remaining after reaction with NaOH:
- Both HNO₃ and NaOH react in a 1:1 molar ratio.
- Initial moles of HNO₃ = 0.01 moles
- Moles of NaOH added = 0.0005 moles
- Moles of HNO₃ remaining = initial moles of HNO₃ - moles of NaOH
[tex]\[
\text{Moles of HNO}_3 \text{ remaining} = 0.01 - 0.0005 = 0.0095 \text{ moles}
\][/tex]
4. Calculate the total volume of the solution after mixing:
- Volume of HNO₃ = 100 ml = 0.1 L
- Volume of NaOH = 10 ml = 0.01 L
- Total volume = 100 ml + 10 ml = 110 ml = 0.11 L
5. Determine the concentration of HNO₃ in the new solution:
- Moles of HNO₃ remaining = 0.0095 moles
- Total volume of solution = 0.11 L
[tex]\[
\text{Concentration of HNO}_3 = \frac{\text{Moles of HNO}_3 \text{ remaining}}{\text{Total volume}}
= \frac{0.0095}{0.11} \approx 0.0864 \text{ M}
\][/tex]
6. Calculate the pH of the solution:
- HNO₃ is a strong acid, it fully dissociates in water.
- pH is calculated as:
[tex]\[
\text{pH} = -\log_{10}(\text{Concentration of HNO}_3)
= -\log_{10}(0.0864) \approx 1.06
\][/tex]
Thus, the pH of the solution after adding 10 ml of 0.05M NaOH to 100 ml of 0.1M HNO₃ is approximately 1.06. Hence, the closest answer choice is:
D. 1.6
