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
