I need help with this exercise 2. Citric acid (C H,O,) is a triprotic acid. Calc
ID: 551586 • Letter: I
Question
I need help with this exercise
2. Citric acid (C H,O,) is a triprotic acid. Calculate the pH and the concentration of all species in a 0.50 M citric acid solution. Kal = 7.59 × 10-4. Kaa = 1.7 × 10-5, Ka,-4.0 × 10-7. a) Using HA to represent citrie acid, write balanced reactions for each of the three dissociation steps. b) Prepare an ICE table for the first dissociation and solve for x. c) Prepare an ICE table for the second dissociation and solve for y d) Prepare an ICE table for the third dissociation and solve for z. e) Calculate [H']. [HjA]. H2A], HA2. A and the pH valueExplanation / Answer
The dissociation equation will be
H3A ---> H+ + H2A-
Initial 0.5 0 0
Change -x +x +x
Equilibrium 0.5-x x x
Ka1 = [H+][H2A-] / [H3A] = x2 / 0.5-x = 7.59 X 10-4
x2= 3.795 X 10-4 - 7.59 X 10-4 x
solving for x , x = 0.0191
(ii) dissociation will be
H2A- ---> H+ + HA-2
Initial 0.0191 0.0191 0
Change -y +y +y
Equilibrium 0.0191-y 0.0191+y y
Ka2 = [H+][HA-] / [H2A-] = y(0.0191+y) / 0.0191-y = 1.7 X 10-5
0.0191y + y2 = 0.03247 X 10-5 - 1.7 X X 10-5 y
y2 + 0.019117y - 0.0000003247 = 0
on solving for y, y = 0.0000169
c) third dissociation will be
HA-2 ---> H+ + A-3
Initial 0.0000169 0.091 0
Change -z +z +z
Equilibrium .0000169 -z 0.091 + z z
Ka3 = [H+][HA-] / [HA-2] = z(0.0191+z) / .0000169-z = 4 X 10-7
we may ignore x in denominator as ka3 is very low
4 X 10-7 = z(0.0191+z) / 0.0000169
4 X 10-7 X 0.0000169 = 0.0191z + z
z = 3.54 X 10-10 M
pH = -log[H+] = -log0.0191 = 1.72