QuestionJuly 31, 2025

What is the pH of a solution made by combining 0.484 m NaC_(2)H_(3)O_(2) with 0.042 M HC_(2)H_(3)O_(2) ? The K_(a) of acetic acid is 1.75times 10^-5 (Report your answer to three decimal places, i.e x.xxx)

What is the pH of a solution made by combining 0.484 m NaC_(2)H_(3)O_(2) with 0.042 M HC_(2)H_(3)O_(2) ? The K_(a) of acetic acid is 1.75times 10^-5 (Report your answer to three decimal places, i.e x.xxx)
What is the pH of a solution made by combining 0.484 m
NaC_(2)H_(3)O_(2) with 0.042 M
HC_(2)H_(3)O_(2) ? The K_(a) of acetic acid is 1.75times 10^-5
(Report your answer to three decimal places, i.e x.xxx)

Solution
4.5(265 votes)

Answer

5.818 Explanation 1. Identify the buffer system The solution contains NaC_{2}H_{3}O_{2} (sodium acetate) and HC_{2}H_{3}O_{2} (acetic acid), forming an acetic acid/acetate buffer. 2. Apply Henderson-Hasselbalch equation Use **pH = pK_a + \log\left(\frac A^- HA \right)**, where [A^-] is the concentration of acetate ion (0.484 \, M) and [HA] is the concentration of acetic acid (0.042 \, M). 3. Calculate pK_a pK_a = -\log(K_a) = -\log(1.75 \times 10^{-5}) = 4.757 4. Calculate pH Substitute values into the Henderson-Hasselbalch equation: pH = 4.757 + \log\left(\frac{0.484}{0.042}\right) 5. Compute logarithm \log\left(\frac{0.484}{0.042}\right) = \log(11.524) \approx 1.061 6. Final calculation pH = 4.757 + 1.061 = 5.818

Explanation

1. Identify the buffer system<br /> The solution contains $NaC_{2}H_{3}O_{2}$ (sodium acetate) and $HC_{2}H_{3}O_{2}$ (acetic acid), forming an acetic acid/acetate buffer.<br /><br />2. Apply Henderson-Hasselbalch equation<br /> Use **pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)**, where $[A^-]$ is the concentration of acetate ion ($0.484 \, M$) and $[HA]$ is the concentration of acetic acid ($0.042 \, M$).<br /><br />3. Calculate pK_a<br /> $pK_a = -\log(K_a) = -\log(1.75 \times 10^{-5}) = 4.757$<br /><br />4. Calculate pH<br /> Substitute values into the Henderson-Hasselbalch equation: <br /> $pH = 4.757 + \log\left(\frac{0.484}{0.042}\right)$<br /><br />5. Compute logarithm<br /> $\log\left(\frac{0.484}{0.042}\right) = \log(11.524) \approx 1.061$<br /><br />6. Final calculation<br /> $pH = 4.757 + 1.061 = 5.818$
Click to rate:

Similar Questions