TITRATION PROBLEM AND SOLVING pH

Question

The problem deals with the titration of 441 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.3 M NaOH.

 

What will the pH of the solution be when 0.2294 L of 1.3 M NaOH are added to the 441 mL of 0.501 M carbonic acid?

Answer 1

Carbonic is a weak diacid. The titration of H₂CO₃ by NaOH shows 2 equivalence points.

The chemical reactions involved in this titration are given by the equation:

 

H₂CO₃ + NaOH → NaHCO₃ + H₂O

NaHCO₃ + NaOH → Na₂CO₃ + H₂O 

Based on the stoichiometry of these reactions: 

At equivalence point 1: mol H₂CO₃ = mol NaOH equivalence 1

M_H2CO3xV_H2CO3 = M_NaOHxV_NaOH^Eq1 

V_NaOH^Eq1= M_H2CO3xV_H2CO3/M_NaOH = 0.501 M x 441 mL/1.3 M = 171.0 mL 

Then: V_NaOH^Eq2= 2xV_NaOH^Eq1= 342.0 mL

The volume of NaOH added is: V_NaOH = 0.2294 L = 229.4 mL

V_NaOH^Eq1<V_NaOH<V_NaOH^Eq2

The pH before equivalence 1 is given by the pH of a buffer solution of H₂CO₃ and HCO₃^-:

pH = pKa₁ + log ([HCO₃^-]/H₂CO₃])

 

The pH at equivalence 1 is: pH = (pK_a1 + pK_a2)/2 = (6.37+10.25)/2 = 8.31

 

The pH between equivalence 1 and equivalence 2 is given by a the pH of a buffer solution of HCO₃^- and CO₃^2-: 

pH = pKa₂ + log ([CO₃^2-]/[HCO₃^-])

 

The pH after equivalence 2 is given by the pH of a strong base: pH = 14 + log[OH^-]_excess

 

[HCO₃^-] = (M_H2CO3xV_H2CO3 - M_NaOHx(V_NaOH -V_NaOH^Eq1))/(V_H2CO3 +V_NaOH) = (441x0.501 - 1.3x(229.4-171.0))/(441+229.4)= 0.2163 M

[CO₃^2-] = M_NaOHx(V_NaOH -V_NaOH^Eq1))/(V_H2CO3 +V_NaOH) = 1.3x(229.4-171.0))/(441+229.4)= 0.1132 M 

 

pH = pKa₂ + log ([CO₃^2-]/[HCO₃^-]) = 10.25 + log (0.1132/0.2163) = 9.969