Question

The next three (3) problems deal with the titration of 411 mL of 0.501 M carbonic acid (H₂CO₃) (K_a1 = 4.3 x 10^-7, K_a2 = 5.6 x 10^-11) with 2.1 M KOH. All answers to three significant digits.

What is the pH of the solution at the 2nd equivalence point?

What will the pH of the solution be when 0.1677 L of 2.1 M KOH are added to the 411 mL of 0.501 M carbonic acid?

How many mL of the 2.1 M KOH are needed to raise the pH of the carbonic acid solution to a pH of 6.714?

Thank you!

Answer 1

H2CO3 +2KOH -----> K2CO3 + 2H2O

from the equation it is clear that each mole of H2CO3 requires 2 moles

volume of KOH required=0.411Lx0.501M/2x0.50/2x2.1M=0.04902L=49.02ml

No of moles of H2CO3 = molarity x volume in L=0.501 Mx0.411L=0.2059

molarity of CO3^2- at second equivalence point=0.2059/(0.411L+0.04902L)=0.4476 M

The equation at second equivalence point is,

HCO3^- +OH^-----> CO3^2- +H2O

the equation involves CO3^2- which is a weak base reacts with water gives back HCO3^-

       CO3^2-+H2O ------> HCO3^- +OH^-

I       0.4476                  0           0

C     -x                           x           x

E     0.4476-x                x           x

Kb= 6.32x10^-5 = [HCO3-][OH-] / [CO3 2-]

6.32x10^-5 = x^2 / 0.4476 - x

x^2 + 6.32x10^-5 - 2.83x10^-5 = 0

x=0.005288 M= [OH-]

pOH = -log [OH-] = 2.28

pH = 14.00 - pOH = 14.00 - 2.28 = 11.72