A 01276g sample of a monoprotic acid molar mass 110 x 102 w

A 0.1276g sample of a monoprotic acid (molar mass = 1.10 x 10^2) was dissolved in 25.0 ml of water and titrated with 0.0633 M NaOH. After 10.0 ml of base had been added, the pH was determined to be 5.47. What is the Ka for the acid?

Solution

step by step solution is here

moles HA = 0.1276/ 1.10 x 10^2 = 0.00116

moles OH- = 0.0100 L x 0.0633 M = 0.000633

HA + OH- >> A- + H2O

moles HA = 0.00116 - 0.000633 = 0.000527
moles A- = 0.000633

total volume = 25 + 10 = 35 mL = 0.0350 L

concentration HA = 0.000527/ 0.0350 L = 0.0151 M
concentration A- = 0.000633 / 0.0350 = 0.0181 M

pH = pKa + log 0.0181 / 0.0151

5.47 = pKa + 0.0787

pKa = 5.39

Ka = 10^- 5.39=4.06 x 10^-6