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

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
Transcript text: In the titration of 38.0 mL of $0.55 \mathrm{M} \mathrm{HF}\left(\mathrm{K}_{\mathrm{a}}=7.2 \times 10^{-4}\right)$ 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
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Solution

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Solution Steps

Step 1: Determine Moles of HF and NaOH

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

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

The moles of NaOH added are:

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

Step 2: Determine Remaining HF and Formed F⁻

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

moles of HF remaining=0.0209mol0.0121mol=0.0088mol \text{moles of HF remaining} = 0.0209 \, \text{mol} - 0.0121 \, \text{mol} = 0.0088 \, \text{mol}

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

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

Step 3: Calculate Concentrations

The total volume after adding NaOH is:

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

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

[HF]=0.0088mol0.0600L=0.1467M [\text{HF}] = \frac{0.0088 \, \text{mol}}{0.0600 \, \text{L}} = 0.1467 \, \text{M}

[F]=0.0121mol0.0600L=0.2017M [\text{F}^-] = \frac{0.0121 \, \text{mol}}{0.0600 \, \text{L}} = 0.2017 \, \text{M}

Step 4: Use the Henderson-Hasselbalch Equation

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

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

First, calculate pKa\text{pK}_a:

pKa=log(7.2×104)=3.1427 \text{pK}_a = -\log(7.2 \times 10^{-4}) = 3.1427

Now, substitute the values into the equation:

pH=3.1427+log(0.20170.1467) \text{pH} = 3.1427 + \log \left( \frac{0.2017}{0.1467} \right)

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

Final Answer

The pH after 22.0 mL of NaOH has been added is approximately:

3.28 \boxed{3.28}

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