Phosphoric acid is a diprotic acid, and a 0.20 M, 50.0 mL sample is titrated wit
ID: 867647 • Letter: P
Question
Phosphoric acid is a diprotic acid, and a 0.20 M, 50.0 mL sample is titrated with 12.5 mL of 0.100 M potassium hydroxide to reach the first equivalence point, where the first acid dissociation constant (ka1) = 7.25 x 10^-3 M at 25 degrees Celsius. An additional 12.5 mL, 0.100 M of potassium hydroxide is added to reach the second equivalence point, where the second acid dissociation constant is (Ka2) = 6.31 x 10^-8 M at 25 degrees Celsius, and the acid is neutralized.
1. Write the balanced chemical equation including al phases of the first dissociation of phosphoric acid.
2. Calculate the PH at the first equivalence point
3. Write the balanced chemical equation including all phases of the second dissociation of phosphoric acid.
4. Calculate the pH at the second equivalence point
Explanation / Answer
H3PO4 + KOH <---> KH2PO4 + H2O
Net inic reaction
1. H3PO4(aq) + OH^- (aq) <---> H2PO4^-(aq) + H2O(l)
Ka1 = [H2PO4^-]/[OH^-][H3PO4]
[OH-] = X
Ka1*[H3PO4]/[H2PO4-]
[H2PO4-] = 0.1*12.5/62.5 = 0.02
[H3PO4]= 0.2*50/62.5 = 0.16
X = 7.25 x 10^-3 *0.16/0.02 = 5.8*10^-2
X = 5.8*10^-2 = [OH-]
pOH = -log 5.8*10^-2 = 2 -log 5.8 = 1.236
pH = 12-1.236 = 10.764
3 .
H3PO4(aq) + OH^- (aq) <---> H2PO4^-(aq) + H2O(l)
H2PO4^-(aq) + OH-(aq) ---> HPO4-(aq) + H2O(L)
ka1*Ka2 = [ H2PO4^-]/[OH^-][H3PO4] * [HPO4-]/[OH-][ H2PO4^-]
ka1*Ka2 = [HPO4-]/ [OH^-]^2[H3PO4]
[HPO4-] = 0.1*12.5/75 = 0.0167
[H3PO4] = 0.2*50/75 = 0.134
6.31 x 10^-8 * 7.25 x 10^-3 = 0.0167/[OH^-]^2*0.134
[OH^-]^2 = 0.0167 /4.574*10^10*0.134
= 2.72*10^-12
[OH-] = 1.65*10^-6
pOH = 6-log 1.65 = 5.78
pH = 14-5.78 = 8.22