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