QuestionAugust 1, 2025

What is the approximate pH at the equivalence point of a weak acid- strong base titration if 25 mL of aqueous HF requires ; 30.00 mL of 0.400 M NaOH? K_(a)HF=6.76times 10^-4 a. 1.74 b. 5.75 c. 8.25 d. 12.26

What is the approximate pH at the equivalence point of a weak acid- strong base titration if 25 mL of aqueous HF requires ; 30.00 mL of 0.400 M NaOH? K_(a)HF=6.76times 10^-4 a. 1.74 b. 5.75 c. 8.25 d. 12.26
What is the approximate pH at the equivalence point of a weak acid- strong base titration if 25 mL of aqueous HF requires ; 30.00 mL of 0.400 M NaOH? K_(a)HF=6.76times 10^-4 a. 1.74 b. 5.75 c. 8.25 d. 12.26

Solution
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Answer

c. 8.25 Explanation 1. Calculate moles of NaOH Moles of NaOH = Volume (L) × Concentration = 0.030 \, \text{L} \times 0.400 \, \text{M} = 0.012 \, \text{mol}. 2. Calculate moles of HF Moles of HF = Volume (L) × Concentration = 0.025 \, \text{L} \times \text{unknown concentration}. At equivalence point, moles of HF = moles of NaOH = 0.012 \, \text{mol}. 3. Determine concentration of F⁻ at equivalence point Total volume = 25 \, \text{mL} + 30 \, \text{mL} = 55 \, \text{mL} = 0.055 \, \text{L}. Concentration of F⁻ = \frac{0.012 \, \text{mol}}{0.055 \, \text{L}} = 0.218 \, \text{M}. 4. Calculate pH using Kb K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.76 \times 10^{-4}} = 1.48 \times 10^{-11}. Use [OH^-] = \sqrt{K_b \cdot [F^-]} = \sqrt{1.48 \times 10^{-11} \cdot 0.218} = 5.64 \times 10^{-6} \, \text{M}. 5. Calculate pOH and then pH pOH = -\log(5.64 \times 10^{-6}) = 5.25. pH = 14 - pOH = 8.75.

Explanation

1. Calculate moles of NaOH<br /> Moles of NaOH = Volume (L) × Concentration = $0.030 \, \text{L} \times 0.400 \, \text{M} = 0.012 \, \text{mol}$.<br />2. Calculate moles of HF<br /> Moles of HF = Volume (L) × Concentration = $0.025 \, \text{L} \times \text{unknown concentration}$. At equivalence point, moles of HF = moles of NaOH = $0.012 \, \text{mol}$.<br />3. Determine concentration of F⁻ at equivalence point<br /> Total volume = $25 \, \text{mL} + 30 \, \text{mL} = 55 \, \text{mL} = 0.055 \, \text{L}$. Concentration of F⁻ = $\frac{0.012 \, \text{mol}}{0.055 \, \text{L}} = 0.218 \, \text{M}$.<br />4. Calculate pH using Kb<br /> $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.76 \times 10^{-4}} = 1.48 \times 10^{-11}$. <br /> Use $[OH^-] = \sqrt{K_b \cdot [F^-]} = \sqrt{1.48 \times 10^{-11} \cdot 0.218} = 5.64 \times 10^{-6} \, \text{M}$.<br />5. Calculate pOH and then pH<br /> $pOH = -\log(5.64 \times 10^{-6}) = 5.25$. <br /> $pH = 14 - pOH = 8.75$.
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