TITRATION Problem Part 2

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. 

 

How many mL of the 1.3 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 5.935?

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: VNaOH = 0.2294 L = 229.4 mL

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 = (pKa1 + pKa2)/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 CO32-:
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

The pH = 5.935 < pH^Eq1 indicating that we do not reach equivalence 1 yet.

 

pH = pKa1 + log ([HCO₃^-]/H₂CO₃])

 

[H₂CO₃] = (M_H2CO3xV_H2CO3 - M_NaOHxVNaOH)/(V_H2CO3 +VNaOH) = (441x0.501 - 1.3xV_NaOH)/(441+229.4)= (220.941-1.3xV_NaOH)/(441+V_NaOH)
[HCO₃^-] = M_NaOHxVNaOH/(V_H2CO3 +V_NaOH) = 1.3xV_NaOH/(441+V_NaOH)

[HCO₃^-]/[CO₃^2-] = (220.941 -1.3xV_NaOH)/1.3xV_NaOH = 10^(pH-pKa1)= 10^(5.935-6.37) = 0.3673 

 

220.941 -1.3xV_NaOH = 0.3673x1.3xV_NaOH 

 

220.941 = 0.3673x1.3xV_NaOH + 1.3xV_NaOH = 1.777xV_NaOH 

 

V_NaOH = 220.941/1.777 = 124.3 mL