Question

A buffer solution based upon ascorbic acid (H2C6H6O6) was created by treating 1.00L of a 1.00 M ascorbic acid solution with NaOH until a pH of 3.602 was achieved (assuming no volume change). To this buffer 1.370 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

For the acid, HC6H6O6 pKa = 4.10

then, we need:

pH = pKa1+ log(A-/HA)

3.602 = 4.10 + log(HA-/H2A)

(A-/HA) = 10^(3.602-4.10) = 0.31768

now..if initially:

mol of acid = MV = 1*1 = 1 mol

then, after "x" mol of NaOH added

mol of acid = 1-x

mol of conjguate = x

(A-/HA) =0.31768

(x)/(1-x) = 0.31768

3.1478x = 1-x

4.1478x = 1

x = 1/4.1478 = 0.2410

that is, we added 0.241 mol of NaOH

....

then, if we add 1.37 mol

mol of NaOH = 1.37 +0.241 = 1.611 mol of NaOH has been added

so

first proton has been neutralized = 1 mol of H2A - 1 mol of OH- = 0 mol of first proton

2nd proton has been partially neutralized = 1-0.611 = 0.389 mol of HA left

mol of NAOH formed conjugate = 0.611

then

A-2 = 0.611

HA- = 0.389

this is 2nd ionization

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

pKA2 = 11.79

pH = 11.79 + log(0.611/0.389 )

pH = 11.986