A student is asked to determine the value of K_a for hypochlorous acid by titrat

Question

A student is asked to determine the value of K_a for hypochlorous acid by titration with sodium hydroxide. The student begins titrating a 36.2 mL sample of a 0.352 M aqueous solution of hypochlorous acid with a 0.257 M aqueous sodium hydroxide solution, but runs out of standardized base before reaching the equivalence point. The student's last observation is that when 15.8 milliliters of sodium hydroxide have been added, the pH is 7.089. What is K_a for hypochlorous acid based on the student's data?

Explanation / Answer

millimoles of acid = 36.2 x 0.352 = 12.74

millimoles of base added = 15.8 x 0.257 = 4.06

12.74 - 4.06 =8.68 millimoles acid left

4.06 moles salt formed

total volume = 15.8 + 36.2 = 52 mL

[salt] = 4.06 / 52 = 0.078 M

[acid] = 8.68 / 52 = 0.167 M

pH = pKa + log [salt] /[acid]

7.089 = pKa + log [0.078] / [0.167]

7.089 = pKa - 0.33

pKa = 7.419

Ka = 10-pKa = 10-7.419

Ka = 3.81 x 10-8

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