Question

A 0.1983 g sample of Na2CO3 was dissolved in 100.00 ml H20. That solution was titrated with 0.1531 M HCL as titrant. The PKa1 and PKa2 of carbonic acid (the conjugate acid of the carbonate ion) are 6.351 and 10.329, respectively.

1.) What is the initial pH of the sodium carbonate solution.

2.) What is the first equivalence point (in ml HCL)? What is the pH at 1st eq.?

3.) What is the second equivalence point (in ml HCL)? What is the pH at 2nd eq?

4.) What is the pH when 18.33 ml of HCL have been added?

Answer 1

[CO₃^2-] =( 0.1983g/105.9884g/mol)×(1000/100.00) = 0.0187096 M

(1)

Initial pH = 7 + 1/2 × (pKa2 + log c) = 7 + 1/2 ×(10.329+log 0.0187096) = 7 + 4.300532

= 11.301 (Answer)

(2)

At first equivalence point , M1V1 = M2V2

0.0187096M×100.00ml = 0.1531M × V₂

V₂ = 12.22 ml (Answer)

A solution of NaHCO₃ is obtained , pH =( pKa1 + pKa2)/2 = (6.351+10.329)/2 = 8.34 (Answer)

(3)

The volume used for second equivalence point = 2 × volume used for first equivalence point = 12.22 ×2 = 24.44 ml. (Answer)

A solution of H₂CO₃ is obtained .

[H₂CO₃] = (0.1983/105.9884)×(1000/124.44) = 0.015035 M

pH = (pKa1 -log c)/2 = (6.351 - log 0.015035)/2

= 4.087 (Answer)

(4)

Millimole of CO₃^2- taken = 0.1983/105.9884 × 1000 = 1.87096mmol

Millimole of HCl added = 18.33ml× 0.1531mol/L = 2.8063 mmol

Final composition of mixture has 0.93536mmol of H₂CO₃ and 0.9356mmol of NaHCO₃ .

It is the middle point between 1st and 2nd equivalence point . (12.22+24.44)/2 = 18.33 ml

pH = pKa1 = 6.351 (Answer)