Question

A buffer solution based upon ascorbic acid (H₂C₆H₆O₆) was created by treating 1.00 L of a 1.00 M ascorbic acid solution with NaOH until a pH of 3.473 was achieved (assuming no volume change). To this buffer 1.300 moles of NaOH were added (assume no volume change). What is the final pH of this solution?
For this problem we can assume the 5% assumption is valid.

Answer 1

let H2C6H6O6 = H2A

H2A <-> H+ + HA-

HA- <-> H+ +A-2

pH = 3.473

initially

pH = pKA + log(HA-/H2A)

pKa1 = 4.17

3.473 = 4.17+ log(HA-/H2A)

(HA-)/(H2A) = 10^(3.473 -4.17) = 0.2009

(HA-)/(H2A) =0.2

HA- + H2A = 1*1 = 1

H2A = 1-HA-

(HA-)/(H2A) =0.2

(HA-) = 0.2*(1-HA-)

(HA-) = 0.2- 0.2HA-

1.2HA- = 0.2

HA- = 0.2/1.2 = 0.16667

H2A = 1-HA- = 1-0.16667 = 0.83333

then

H2A = 0.83333

HA- = 0.16667

then... after addition of 1.3 mol of OH-

H2A + OH- <-- > HA-

H2A = 0.83333 - 0.83333 = 0

HA- = 0.16667 + 0.83333 = 1

OH- left = 1.3 -0.83333 = 0.46667

now..

HA- + OH- <-> A-2 + H2O

HA- = 1-0.46667= 0.53333

A-2 left = 0.46667

this is 2nd ionization

pH = pKa2 + log(A-2/HA-)

pH = 11.57+ log(0.46667/0.53333)

pH = 11.512