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 34.2 mL sample of a 0.459 M aqueous solution of hypochlorous acid with a 0.453 M aqueous sodium hydroxide solution, but runs out of standardized base before reaching the equivalence point. The student's last observation is that when 21.9 milliliters of sodium hydroxide have been added, the pH is 7.665.

What is K_a for hypochlorous acid based on the student's data?

Answer 1

Given moles of HClO = (0.459mol/1000ml) × 34.2ml = 0.01570mol

moles of NaOH consumed = (0.453mol/1000ml)× 21.9ml = 0.0099207mol

The reaction between NaOH and HClO is 1:1molar

HClO + OH^- ------> H2O + ClO^-

0.0099207moles of HClO moles of OH^- react with 0.0099207moles of OH^- to form 0.0099207moles of OCl^-

After addition of NaOH

moles of HClO = 0.01570mol - 0.0099207mol = 0.0057793mol

moles of ClO^- = 0.0099207mol

total volume = 34.2ml + 21.9ml = 56.1ml

[HClO] = (0.0057793mol/56.1ml)×1000ml = 0.1030M

[ClO^-] = (0.0099207mol/56.1ml)×1000ml = 0.1768M

Henderson- Hasselbalch equation is

pH = pKa + log([A^- ]/[HA])

7.665 = pKa + log(0.1768M/0.1030M)

7.665 = pKa + 0.235

pKa = 7.43

pKa = -logKa

-logKa= 7.43

Ka = 3.72×10^-8