Question 8.4: Using the Henderson-Hasselbalch Equation. Sodium hypochlorit...
Using the Henderson-Hasselbalch Equation
Sodium hypochlorite (NaOCl, the active ingredient of almost all bleaches) was dissolved in a solution buffered to pH 6.20. Find the ratio [OCl^{−}]/[HOCl] in this solution.
In Appendix G, we find that pK_{a} = 7.53 for hypochlorous acid, HOCl. The pH is known, so the ratio [OCl^{−}]/[HOCl] can be calculated from the Henderson-Hasselbalch equation.
HOCl \xrightleftharpoons[]{} H^{+} + OCl^{−}
pH = pK_{a} + \log \frac{[OCl^{−}]}{[HOCl]}
6.20 = 7.53 + \log \frac{[OCl^{−}]}{[HOCl]}
− 1.33 = \log \frac{[OCl^{−}]}{[HOCl]}
10^{−1.33} = 10^{\log ([OCl^{−}]/[HOCl])} = \frac{[OCl^{−}]}{[HOCl]}
0.047 = \frac{[OCl^{−}]}{[HOCl]}
The ratio [OCl^{−}]/[HOCl] is set by pH and pK_{a}. We do not need to know how much NaOCl was added, or the volume.
Test Yourself Find [OCl^{−}]/[HOCl] if pH is raised by one unit to 7.20. (Answer: 0.47)
10^{\log z} = z