Question

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

Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.5 M H3PO3(aq) with 1.5 M KOH(aq).

pKa1= 1.30

pKa2= 6.70

(a) before addition of any KOH =_____number

(b) after addition of 25.0 mL of KOH = ______number

(c) after addition of 50.0 mL of KOH = ______number

(d) after addition of 75.0 mL of KOH =_______number

(e) after addition of 100.0 mL of KOH=______number

please explain how you got each! thanks so much

Answer 1

H3PO3 millimoles = 50 x 1.5 = 75

a) Before any addition of KOH :

H3PO3 ---------------------> H+ +   H2PO3-

1.5 0             0

1.5- x                                   x              x

Ka1 = [H+][H2PO3-] / [H3PO3]

0.05 = x^2 / 1.5 - x

x = 0.25

[H+] = 0.25 M

pH = -log [H+] = -log [0.25]

       = 0.6

pH = 0.6

b) after addition of 25.0 mL KOH

it is first equivaelce point

pH = pKa1 = 1.30

pH = 1.30

c ) addition of 50.0 mL KOH

it is first equivalence point

pH = 1/2 (pKa1 + pKa2)

pH =1/2 (1.30 + 6.70)

pH = 4.0

d) 75.0 mL KOH

it is seond half equivalece point

pH = pKa2

pH = 6.70

4) 100.0 mL KOH

it is second equivalece point

HPO3^-2 millimoles = 100 x 1.5 = 150

HPO3^-2 molarity = 150 / (50 +100) = 1.0 M

HPO3^-2 + H2O ------------------> H2PO4- + OH-

1.0 -x                                           x                x

Kb2 = x^2 / 1.0-x

5.01 x 10^-8 = x^2 / 1.0 -x

x^2 + 5.01 x 10^-8 - 5.01 x 10^-8 = 0

x = 2.24 x 10^-4

[OH-] = 2.24 x 10^-4 M

pOH = -log[OH-] = -log (2.65 x 10^-4 )

pOH = 3.65

pH + pOH = 14

pH = 10.35