Question

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture.

pKa1 = 1.30

pKa2 = 6.70

Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq).

A) before addition of any KOH

B) after addition of 25.0 mL of KOH

C) after addition of 50.0 mL of KOH

D) after addition of 75.0 mL of KOH

E) after addition of 100.0 mL of KOH

Answer 1

[H3PO3] = 2.7 M

Volume of H3PO3 = 50 mL = 0.050 L

pKa1 = 1.30 , Ka1 = 0.050

pKa2 = 6.70 , Ka2 = 1.99x10^-7

A). before addition of any KOH

H3PO3      H+ + H2PO3-

initially 2.7 0 0

finally 2.7 - x x x

Ka1 = x*x / 2.7 - x

0.050 = x² / 2.7 - x

0.135 - 0.050*x = x²

x² + 0.050*x - 0.135 = 0

x = 0.343

[H2PO3-] = [H+] = 0.343 M

Similarly,

H2PO3-    H+ + HPO3^2-

initially 0.343 x x

finally 0.343 - x x x

Ka2 = x*x / 0.343 - x

1.99x10^-7 = x² / 0.343 - x

x = 0.00026

[H+] = [HPO3^2-] = 0.00026 M

total [H+] = 0.343 + 0.00026

= 0.34326M

pH = - log 0.34326

= 0.464

B) after addition of 25.0 mL of KOH

[H+] = 0.34326M

moles of H+ = 0.34326*0.050

= 0.017163

[KOH] = 2.7 M

moles of OH- = 2.7*0.025

= 0.0675

H3PO3 + 2KOH    K2HPO3 + 2H2O

moles of K2HPO3 formed = 0.017163

[K2HPO3] = 0.017163 / 0.075 = 0.2288 M

moles of OH- left = 0.0675 - 2*0.017163 = 0.033174

[OH-] = 0.44232 M

pH = pKa + log [OH-] / [K2HPO3]

= 1.30 + log (0.44232 / 0.2288)

= 1.30 + 0.286

= 1.586

C) after addition of 50.0 mL of KOH

moles of OH- = 0.135

moles of OH- left = 0.135 - 2*0.017163

= 0.100674

[OH-] = 1.3423 M

pH = pKa + log [OH-] / [K2HPO3]

= 1.30 + log (1.3423 / 0.2288)

= 1.30 + 0.768

= 1.586