Question

Phosphorous acid, H, PO, (aq), is a diprotic oxyacid that is an important compound in industry and agriculture. pKalp Ka2 1.30 6.70 но —р —он Calculate the pH for each of the points in the titration of 50.0 mL of 2.1 MH,PO, (aq) with 2.1 M KOH(aq). before addition of any KOH: after addition of 25.0 mL KOH: after addition of 50.0 mL KOH: after addition of 75.0 mL KOH: after addition of 100.0 mL KOH:

Answer 1

1) before addition of any KOH :

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

2.1                                     0             0

2.1- x                                   x              x

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

0.05 = x^2 / 2.1 - x

x^2 + 0.05 x - 0.105 = 0

x = 0.30

[H+] = 0.30 M

pH = -log [H+] = -log [0.30]

       = 0.523

pH = 0.523

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

e) 100.0 mL KOH

it is second equivalece point

HPO3^-2 millimoles = 100 x 2.1 = 210

HPO3^-2 molarity = 210 / (50 +100) = 1.4M

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

1.4 -x                                           x                x

Kb2 = x^2 / 1.4-x

5.01 x 10^-8 = x^2 / 1.4-x

x^2 + 5.01 x 10^-8 - 7.017 x 10^-8 = 0

x = 2.65 x 10^-4

[OH-] = 2.65 x 10^-4 M

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

pOH = 3.58

pH + pOH = 14

pH = 10.42