A student is asked to determine the value of Ka for
nitrous acid by titration with potassium
hydroxide.
The student begins titrating a 48.3 mL sample of a
0.360 M aqueous solution of nitrous
acid with a 0.216 M aqueous
potassium hydroxide solution, but runs out of
standardized base before reaching the equivalence point. The
student's last observation is that when 53.2
milliliters of potassium hydroxide have been
added, the pH is 3.603.
What is Ka for nitrous acid based on
the student's data?
The millimoles of nitrous acid = 48.3 x 0.360 = 17.388
The millimoles of KOH added = 53.2 x 0.216 = 11.491
17.388 - 11.491 = 5.897 millimoles acid left
11.491 millimoles salt formed
Total volume = 48.3 + 53.2 = 101.5 mL
[acid] = 5.897 / 101.5 = 0.0580
[salt] = 11.491 / 101.5 = 0.1132
Now,
pH = pKa + log [salt] / [acid]
3.603 = pKa + log [0.1132 / 0.0580]
3.603 = pKa + log (1.951)
3.603 = pKa + 0.2902
pKa = 3.603 - 0.2902
pKa = 3.3128
Ka = 10-3.3128
Ka = 0.000486
= 4.86 * 10-4
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