Questions: Which buffer solution has the lowest pH (HF has K2=7.2 × 10^-4)? - 0.10 M HF and 0.10 M NaF - 0.20 M HF and 0.20 M NaF - 0.20 M HF and 0.10 M NaF - 0.10 M HF and 0.20 M NaF - All of the buffer solutions described have the same pH.

Which buffer solution has the lowest pH (HF has K2=7.2 × 10^-4)?
- 0.10 M HF and 0.10 M NaF
- 0.20 M HF and 0.20 M NaF
- 0.20 M HF and 0.10 M NaF
- 0.10 M HF and 0.20 M NaF
- All of the buffer solutions described have the same pH.

Transcript text: Which buffer solution has the lowest pH (HF has $\mathrm{K}_{2}=7.2 \times 10^{-4}$ )? - 0.10 M HF and 0.10 M NaF - 0.20 M HF and 0.20 M NaF - 0.20 M HF and 0.10 M NaF - 0.10 M HF and 0.20 M NaF - All of the buffer solutions described have the same pH.

Solution

Solution Steps

Step 1: Identify the Henderson-Hasselbalch Equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where:

$\text{p}K_a = -\log K_a$
$[\text{A}^-]$ is the concentration of the conjugate base (NaF)
$[\text{HA}]$ is the concentration of the weak acid (HF)

Step 2: Calculate $\text{p}K_a$ for HF

Given $K_a = 7.2 \times 10^{-4}$: \[ \text{p}K_a = -\log(7.2 \times 10^{-4}) \approx 3.1427 \]

Step 3: Apply the Henderson-Hasselbalch Equation to Each Buffer Solution

0.10 M HF and 0.10 M NaF: \[ \text{pH} = 3.1427 + \log \left( \frac{0.10}{0.10} \right) = 3.1427 + \log(1) = 3.1427 \]
0.20 M HF and 0.20 M NaF: \[ \text{pH} = 3.1427 + \log \left( \frac{0.20}{0.20} \right) = 3.1427 + \log(1) = 3.1427 \]
0.20 M HF and 0.10 M NaF: \[ \text{pH} = 3.1427 + \log \left( \frac{0.10}{0.20} \right) = 3.1427 + \log(0.5) = 3.1427 - 0.3010 = 2.8417 \]
0.10 M HF and 0.20 M NaF: \[ \text{pH} = 3.1427 + \log \left( \frac{0.20}{0.10} \right) = 3.1427 + \log(2) = 3.1427 + 0.3010 = 3.4437 \]

Step 4: Compare the pH Values