Question

A volume of 500.0 mL of 0.170 M NaOH is added to 595 mL of 0.200 M weak acid (𝐾a=6.74×10−5). What is the pH of the resulting buffer?

HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)

Answer 1

mols HA = M x L = 0.200 x 0.595 = 0.119
mols NaOH = 0.170 x 0.500 = 0.085
So you form 0.085 mols NaA and you have remaining 0.119 - 0.085 = 0.034 mols HA.
(acid) = mols/L = 0.034/(0.500 + 0.595) = 0.034/1.095 = ?
(base) = the salt = 0.085/1.095 = ?
pH = pKa + log [(base)/acid)]
Substitute the data and calculate the pH of the solution. Post your work if you get stuck.