Question

You want a buffer with a pH of 7.34. How much 0.10 M NaOH would you have to add to 75 mL of 0.10 M hypochlorous acid to make this buffer?

Answer 1

First, find pKa of HClO

pKa = 7.5

so...

if we want a pH = 7.34

apply buffer equation:

pH = pKa + log([ClO-]/[HClO])

so

7.34 = 7.5 + log([ClO-]/[HClO])

ratio = 10^(7.34-7.5) = 0.6918

[ClO-]/[HClO] = 0.6918

[ClO-] = 0.6918 * [HClO]

so..

initially...

we have only HClO:

mmol = MV = 0.1*75 = 7.5 mmol of HClO

mmol of Cl-O = 0

after additoin of

mmol of base = M*V = 0.1*Vbase

then

mmmol of acid left = 7.5 - 0.1*Vbase

mmol of ClO- formed = 0 + 0.1*Vbase

so..

[ClO-] = 0.6918 * [HClO]

0.1*Vbase = 0.6918 * (7.5 - 0.1*Vbase)

0.1/0.6918 *Vbase = 7.5 - 0.1*Vbase

(0.14455+0.1)Vbase = 7.5

Vbase = 7.5/(0.14455+0.1) = 30.668 mL of base required