The pKa of HF is 3.17. a. 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 40.0 ml of 10.0 M HCl?
The correct answer and explanation is:
Let’s solve this step-by-step.
Given:
- pKapK_a of HF = 3.17
- Initial solution: 1.00 L of 1.00 M HF and 1.50 M NaF
- Volume of HCl added = 40.0 mL (0.0400 L) of 10.0 M HCl
Part (a): Calculate the pH of the buffer solution
This is a classic buffer problem containing a weak acid (HF) and its conjugate base (F⁻ from NaF). The pH can be found using the Henderson-Hasselbalch equation: pH=pKa+log([A−][HA])pH = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
- [HA][\text{HA}] = concentration of HF = 1.00 M
- [A−][\text{A}^-] = concentration of F⁻ from NaF = 1.50 M
Substitute values: pH=3.17+log(1.501.00)pH = 3.17 + \log\left(\frac{1.50}{1.00}\right)
Calculate the logarithm: log(1.50)=0.176\log(1.50) = 0.176
Therefore: pH=3.17+0.176=3.35pH = 3.17 + 0.176 = 3.35
Part (b): Calculate the pH after addition of 40.0 mL of 10.0 M HCl
First, calculate moles of species before HCl addition:
- Moles HF initially = 1.00 mol/L×1.00 L=1.00 mol1.00 \, \text{mol/L} \times 1.00 \, \text{L} = 1.00 \, \text{mol}
- Moles F⁻ initially = 1.50 mol/L×1.00 L=1.50 mol1.50 \, \text{mol/L} \times 1.00 \, \text{L} = 1.50 \, \text{mol}
Moles of HCl added: 0.0400 L×10.0 mol/L=0.400 mol HCl0.0400 \, \text{L} \times 10.0 \, \text{mol/L} = 0.400 \, \text{mol HCl}
HCl is a strong acid and will react with the base (F⁻) to form HF: F−+H+→HF\text{F}^- + \text{H}^+ \rightarrow \text{HF}
Calculate new moles after reaction:
- New moles F⁻ = 1.50−0.400=1.10 mol1.50 – 0.400 = 1.10 \, \text{mol}
- New moles HF = 1.00+0.400=1.40 mol1.00 + 0.400 = 1.40 \, \text{mol}
The total volume after adding HCl is: 1.00 L+0.0400 L=1.040 L1.00 \, \text{L} + 0.0400 \, \text{L} = 1.040 \, \text{L}
Calculate new concentrations: [F−]=1.101.040=1.058 M[\text{F}^-] = \frac{1.10}{1.040} = 1.058 \, M [HF]=1.401.040=1.346 M[\text{HF}] = \frac{1.40}{1.040} = 1.346 \, M
Use Henderson-Hasselbalch equation again: pH=3.17+log(1.0581.346)pH = 3.17 + \log\left(\frac{1.058}{1.346}\right)
Calculate the ratio and logarithm: 1.0581.346=0.786\frac{1.058}{1.346} = 0.786 log(0.786)=−0.104\log(0.786) = -0.104
Thus: pH=3.17−0.104=3.07pH = 3.17 – 0.104 = 3.07
Summary:
- Initial pH = 3.35
- pH after HCl addition = 3.07
Explanation:
The initial solution is a buffer consisting of a weak acid (HF) and its conjugate base (F⁻). The Henderson-Hasselbalch equation relates the pH to the ratio of conjugate base to acid. Adding HCl introduces strong acid, which reacts with the base F⁻, reducing its amount and increasing the HF concentration. This shifts the ratio in favor of the acid, lowering the pH. The buffer resists large pH changes, so although a significant amount of HCl is added, the pH changes only moderately from 3.35 to 3.07. The slight volume increase is accounted for in concentration calculations, which slightly affects the final pH. This example shows the effectiveness of buffers in maintaining relatively stable pH when small amounts of strong acid are added.