What volume of 2.00 m must be added to 150.0 ml of 1.00 m chloroacetic acid to produce a buffer solution having a ph of 3.30?

PH = pKa + log{[A-]/[HA]}

4.00 = 3.83 + log{[A-]/[HA]}

[A-]/[HA] = 10^ (4.00 - 3.83) = 10^ (0.17) = 1.48

At this pH, the number of moles of base and acid will be:

A - = x

HA = 0.3 - x

Substituting these values into the equation above gives:

x / (0.3 - x) = 1.48

Multiplying both sides by (0.3 - x) and bringing like-terms to the same side yields:

2.48x = 0.44

x = 0.18

This means that we require 0.18 moles of NaOH or

(0.18 moles) x [ (1000 mL) / (2 moles NaOH) ] = 90 mL NaOH