QuestionAugust 4, 2025

An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of 0.150 M HClO and 0.100 M NaCIO. Be sure your answer has the correct number of significant figures. Note: Reference the K_(a) of acids at 25^circ C table for additional information. pH=square

An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of 0.150 M HClO and 0.100 M NaCIO. Be sure your answer has the correct number of significant figures. Note: Reference the K_(a) of acids at 25^circ C table for additional information. pH=square
An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of 0.150 M HClO and 0.100 M NaCIO. Be sure your answer has the correct number of significant figures. Note: Reference the K_(a) of acids at 25^circ C table for additional information. pH=square

Solution
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Answer

pH = 7.36 Explanation 1. Identify the reaction The solution is a buffer containing HClO and NaClO. The equilibrium involves HClO \rightleftharpoons ClO^- + H^+. 2. Use Henderson-Hasselbalch equation **pH = pK_a + \log\left(\frac A^- HA \right)** where [A^-] = 0.100 \, M (NaClO) and [HA] = 0.150 \, M (HClO). 3. Find pK_a Given K_a = 2.9 \times 10^{-8} for HClO, calculate pK_a: pK_a = -\log(2.9 \times 10^{-8}) = 7.54. 4. Calculate pH Substitute values into the Henderson-Hasselbalch equation: pH = 7.54 + \log\left(\frac{0.100}{0.150}\right). 5. Compute logarithm \log\left(\frac{0.100}{0.150}\right) = \log(0.6667) = -0.176. 6. Final calculation pH = 7.54 - 0.176 = 7.36.

Explanation

1. Identify the reaction<br /> The solution is a buffer containing HClO and NaClO. The equilibrium involves $HClO \rightleftharpoons ClO^- + H^+$.<br /><br />2. Use Henderson-Hasselbalch equation<br /> **$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$** where $[A^-] = 0.100 \, M$ (NaClO) and $[HA] = 0.150 \, M$ (HClO).<br /><br />3. Find $pK_a$<br /> Given $K_a = 2.9 \times 10^{-8}$ for HClO, calculate $pK_a$: $pK_a = -\log(2.9 \times 10^{-8}) = 7.54$.<br /><br />4. Calculate pH<br /> Substitute values into the Henderson-Hasselbalch equation: $pH = 7.54 + \log\left(\frac{0.100}{0.150}\right)$.<br /><br />5. Compute logarithm<br /> $\log\left(\frac{0.100}{0.150}\right) = \log(0.6667) = -0.176$.<br /><br />6. Final calculation<br /> $pH = 7.54 - 0.176 = 7.36$.
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