Question

What is the pH of a solution after the addition of 0.0075 mol of solid NaOH to 100 mL of a 0.10 M solution of acetic acid? The pka of acetic acid is 4.7.

Answer 1

volume of acetic acid solution = 100 mL
= 0.1 L
number of mole of acetic acid = (molarity)*(volume in L)
= 0.1*0.1
= 0.01 mol

number of mole of NaOH = 0.0075 mol
NaOH react with acetic acid to form salt
number of mol of salt formed = 0.0075 mole
number of mole of acetic acid remaining = 0.01 - 0.0075
= (2.5*10^-3) mol
= 0.0025 mol

it forms buffer solution
use,
pH = pKa + log([salt]/[acid])
pH = 4.7 + log(0.0075/0.0025)
pH = 4.7+ log(3)
pH = 4.7 + 0.477
pH = 5.177

Answer : 5.177