a) The pKa of HF is 3.17. Calculate the pH of a 1.00 L solution that is 1.00 M HF and 1.50 M NaF. b) What is the pH of this solution after addition of 50.0 mL of 10.0 M HCl?
The Correct Answer and Explanation is :
Part A: pH of a 1.00 M HF and 1.50 M NaF Solution
The solution contains both a weak acid (HF) and its conjugate base (F⁻) from NaF. This setup is a buffer solution, which we can analyze using the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
Where:
- pKa of HF = 3.17
- [A⁻] = concentration of the conjugate base (F⁻) = 1.50 M (from NaF)
- [HA] = concentration of the weak acid (HF) = 1.00 M
Substituting the values into the equation:
[
\text{pH} = 3.17 + \log \left( \frac{1.50}{1.00} \right)
]
First, calculate the ratio of concentrations:
[
\frac{1.50}{1.00} = 1.50
]
Now calculate the logarithm:
[
\log(1.50) \approx 0.176
]
Therefore, the pH is:
[
\text{pH} = 3.17 + 0.176 = 3.35
]
Thus, the pH of the solution is 3.35.
Part B: pH After Addition of 50.0 mL of 10.0 M HCl
After adding HCl, the solution will experience an increase in the concentration of ( \text{H}^+ ) ions, which will interact with the conjugate base ( \text{F}^- ) in the buffer. The reaction will be:
[
\text{F}^- + \text{H}^+ \rightleftharpoons \text{HF}
]
First, calculate the number of moles of HCl added:
[
\text{moles of HCl} = (10.0 \, \text{M}) \times (50.0 \, \text{mL}) = 10.0 \times 0.050 = 0.50 \, \text{mol}
]
The initial total volume is 1.00 L of buffer solution + 0.050 L of HCl solution, so the final volume is:
[
\text{final volume} = 1.00 \, \text{L} + 0.050 \, \text{L} = 1.05 \, \text{L}
]
Next, calculate the change in the concentrations of F⁻ and HF:
- Initial moles of F⁻ = ( 1.50 \, \text{M} \times 1.00 \, \text{L} = 1.50 \, \text{mol} )
- Initial moles of HF = ( 1.00 \, \text{M} \times 1.00 \, \text{L} = 1.00 \, \text{mol} )
After the addition of HCl, the ( \text{H}^+ ) ions will react with F⁻, converting some F⁻ to HF. The moles of F⁻ will decrease, and the moles of HF will increase by the amount of H⁺ added. Since 0.50 mol of HCl is added, this will react with an equivalent amount of F⁻.
After the reaction:
- Moles of F⁻ = ( 1.50 – 0.50 = 1.00 \, \text{mol} )
- Moles of HF = ( 1.00 + 0.50 = 1.50 \, \text{mol} )
Now, calculate the new concentrations:
- [F⁻] = ( \frac{1.00 \, \text{mol}}{1.05 \, \text{L}} = 0.952 \, \text{M} )
- [HF] = ( \frac{1.50 \, \text{mol}}{1.05 \, \text{L}} = 1.43 \, \text{M} )
Using the Henderson-Hasselbalch equation again:
[
\text{pH} = 3.17 + \log \left( \frac{0.952}{1.43} \right)
]
First, calculate the ratio:
[
\frac{0.952}{1.43} \approx 0.665
]
Now calculate the logarithm:
[
\log(0.665) \approx -0.176
]
Thus, the pH is:
[
\text{pH} = 3.17 + (-0.176) = 2.99
]
Therefore, the pH of the solution after the addition of HCl is 2.99.
Summary of Results:
- pH of the buffer solution (1.00 M HF, 1.50 M NaF) = 3.35
- pH after addition of 50.0 mL of 10.0 M HCl = 2.99