Consider the titration of 100.00 mL of 1.00 M H_3PO_4 with 1.00 M NaOH .K_a1 = 7
ID: 508803 • Letter: C
Question
Consider the titration of 100.00 mL of 1.00 M H_3PO_4 with 1.00 M NaOH .K_a1 = 7.5 times 10^-3, K_a2 = 6.2 times 10^-8, K_a3 = 4.2 times 10^-13 a. Calculate the pH before any base is added. b. Calculate the pH after 25.00 mL of base is added. c. Calculate pH after 50.00 mL base is added. d. Calculate pH after 100.00 mL base is added. e. Calculate pH after 150.00 mL base is added. f. Calculate pH after 200.00 mL base is added. g. Calculate pH after 225 .00 mL base is added. h. Calculate pH after 250.00 mL base is added. i. Calculate pH after 300.00 mL base is added. j. Calculate pH after 350.00 mL base is added.Explanation / Answer
Phosphoric acid, H3PO4 , is triprotic, so there are three equilibria to consider:
H3PO4 (aq) + H2O (l) = H2PO4- (aq) + H3O+ (aq)
Ka1 = 7.5 x10-3 .....................(i)
H2PO4- (aq) + H2O (l) = HPO42- (aq) + H3O+ (aq)
Ka2 = 6.2 x10-8 .....................(ii)
HPO42- (aq) + H2O (l) = PO43- (aq) + H3O+ (aq)
Ka3 = 4.2 x10-13 ...................(iii)
Each step has a separate Ka.
Ka1 >> > Ka2 and Ka3.
which says that majority of Hydronium is produced during step one so we can ignore step (ii) and (iii).
Since H3PO4 is weak acid it will not dissociate completely and we have to write the ICE.
H3PO4 (aq) + H2O (l) = H2PO4- (aq) + H3O+ (aq)
I 1 0 0
C -x +x +x
E 1-x x x
Ka1 = 7.5x10-3 = [H3O+][H2PO4-]/[H3PO4] = (x)(x)/(1-x)
Consider x <<< 1, (1-x) = 1
0.75 x 10-4 = x2
[H3O+] = x = 0.866 x 10-2
pH = log (1/[H3O+]) = log (100/0.866) = log (115.47)
pH of acid before any base is added = 2.062