QuestionMay 2, 2025

Question 11 (1 point) What is the pH of a buffer that is 0.451 M HF and 0.118 M LIF?The K_(a) for HF is 35 times 10^-4 Your Answer: square

Question 11 (1 point) What is the pH of a buffer that is 0.451 M HF and 0.118 M LIF?The K_(a) for HF is 35 times 10^-4 Your Answer: square
Question 11 (1 point)
What is the pH of a buffer that is 0.451 M HF and 0.118 M LIF?The K_(a) for HF is 35
times 10^-4
Your Answer:
square

Solution
3.6(207 votes)

Answer

2.878 Explanation 1. Identify the formula Use the Henderson-Hasselbalch equation: **pH = pK_a + \log\left(\frac A^- HA \right)**. 2. Calculate pK_a pK_a = -\log(K_a) = -\log(3.5 \times 10^{-4}) \approx 3.46. 3. Apply concentrations to the formula Substitute [A^-] = 0.118 \, M and [HA] = 0.451 \, M into the equation: pH = 3.46 + \log\left(\frac{0.118}{0.451}\right). 4. Calculate the logarithmic term \log\left(\frac{0.118}{0.451}\right) \approx \log(0.2616) \approx -0.582. 5. Calculate the final pH pH = 3.46 - 0.582 = 2.878.

Explanation

1. Identify the formula<br /> Use the Henderson-Hasselbalch equation: **$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$**.<br /><br />2. Calculate $pK_a$<br /> $pK_a = -\log(K_a) = -\log(3.5 \times 10^{-4}) \approx 3.46$.<br /><br />3. Apply concentrations to the formula<br /> Substitute $[A^-] = 0.118 \, M$ and $[HA] = 0.451 \, M$ into the equation: <br /> $pH = 3.46 + \log\left(\frac{0.118}{0.451}\right)$.<br /><br />4. Calculate the logarithmic term<br /> $\log\left(\frac{0.118}{0.451}\right) \approx \log(0.2616) \approx -0.582$.<br /><br />5. Calculate the final pH<br /> $pH = 3.46 - 0.582 = 2.878$.
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