Citric acid (H 3 C 6 H 5 O 7 , K a 1 = 7.4 x 10 -4 , K a 2 = 1.7 x 10 - 5 , and
ID: 828268 • Letter: C
Question
Citric acid (H3C6H5O7, Ka1 = 7.4 x 10-4, Ka2 = 1.7 x 10-5, and Ka3 = 4.0 x 10-7) is titrated with a NaOH solution. When the pH reaches 5.80, what are the main species in the solution?
Answer is HC6H5O72- and C6H5O73-, [HC6H5O72-] > [C6H5O73-]
Please expain why this is the answer?
H3C6H5O7 and H2C6H5O7-, [H2C6H5O7-] > [H3C6H5O7] H2C6H5O7- and HC6H5O72-, [H2C6H5O7-] > [HC6H5O72-] HC6H5O72- and C6H5O73-, [HC6H5O72-] > [C6H5O73-] HC6H5O72- and C6H5O73-, [HC6H5O72-] = [C6H5O73-] HC6H5O72-onlyExplanation / Answer
pH = pKa1 + log([H2C6H5O7-]/[H3C6H5O7]) = -log Ka1 + log([H2C6H5O7-]/[H3C6H5O7])
5.80 = -log(7.4 x 10^(-4)) + log([H2C6H5O7-]/[H3C6H5O7])
log([H2C6H5O7-]/[H3C6H5O7]) = 2.669
[H2C6H5O7-]/[H3C6H5O7] = 10^2.669 = 467
=> [H2C6H5O7-] = 467 x [H3C6H5O7]
=> [H2C6H5O7-] > [H3C6H5O7]
pH = pKa2 + log([HC6H5O72-]/[H2C6H5O7-]) = -log Ka2 + log([HC6H5O72-]/[H2C6H5O7-])
5.80 = -log(1.7 x 10^(-5)) + log([HC6H5O72-]/[H2C6H5O7-])
log([HC6H5O72-]/[H2C6H5O7-]) = 1.0304
[HC6H5O72-]/[H2C6H5O7-] = 10^1.0304 = 10.7
=> [HC6H5O72-] = 10.7 x [H2C6H5O7-]
=> [HC6H5O72-] > [H2C6H5O7-]
pH = pKa3 + log([C6H5O73-]/[HC6H5O72-]) = -log Ka3 + log([C6H5O73-]/[HC6H5O72-])
5.80 = -log(4.0 x 10^(-7)) + log([C6H5O73-]/[HC6H5O72-])
log([C6H5O73-]/[HC6H5O72-]) = -0.59794
[C6H5O73-]/[HC6H5O72-] = 10^(-0.59794) = 0.252
=> [C6H5O73-] = 0.252 x [HC6H5O72-]
=> [HC6H5O72-] = 3.97 x [C6H5O73-]
=> [HC6H5O72-] > [C6H5O73-]
Comparing the relative concentrations above gives:
[HC6H5O72-] > [C6H5O73-] > [H2C6H5O7-] > [H3C6H5O7]
The main species are:
HC6H5O72- and C6H5O73-, [HC6H5O72-] > [C6H5O73-]