Questions: In the titration of 38.0 mL of 0.55 M HF(Ka=7.2 x 10^-4) with 0.55 M NaOH, what is the pH after 22.0 mL of NaOH has been added? (A) 3.28 (B) 3.14 (C) 1.99 (D) 9.67 (E) 12.96

Solution Steps

First, calculate the moles of HF and NaOH. The initial moles of HF are given by:

$$\text{moles of HF} = 0.0380 \, \text{L} \times 0.55 \, \text{M} = 0.0209 \, \text{mol}$$

The moles of NaOH added are:

$$\text{moles of NaOH} = 0.0220 \, \text{L} \times 0.55 \, \text{M} = 0.0121 \, \text{mol}$$

Since NaOH reacts with HF in a 1:1 ratio, the moles of HF remaining after reaction are:

$$\text{moles of HF remaining} = 0.0209 \, \text{mol} - 0.0121 \, \text{mol} = 0.0088 \, \text{mol}$$

The moles of fluoride ion ($\text{F}^-$) formed are equal to the moles of NaOH added:

$$\text{moles of F}^- = 0.0121 \, \text{mol}$$

The total volume after adding NaOH is:

$$\text{total volume} = 0.0380 \, \text{L} + 0.0220 \, \text{L} = 0.0600 \, \text{L}$$

The concentrations of HF and $\text{F}^-$ are:

$$[\text{HF}] = \frac{0.0088 \, \text{mol}}{0.0600 \, \text{L}} = 0.1467 \, \text{M}$$

$$[\text{F}^-] = \frac{0.0121 \, \text{mol}}{0.0600 \, \text{L}} = 0.2017 \, \text{M}$$

The pH can be calculated using the Henderson-Hasselbalch equation:

$$\text{pH} = \text{pK}_a + \log \left( \frac{[\text{F}^-]}{[\text{HF}]} \right)$$

First, calculate $\text{pK}_a$:

$$\text{pK}_a = -\log(7.2 \times 10^{-4}) = 3.1427$$

Now, substitute the values into the equation:

$$\text{pH} = 3.1427 + \log \left( \frac{0.2017}{0.1467} \right)$$

$$\text{pH} = 3.1427 + \log(1.3751) = 3.1427 + 0.1378 = 3.2805$$

Final Answer