Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound
ID: 494337 • Letter: P
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).
McQuarrie Rock Gallogly presented by Sapling Learning Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is pKa1 pKa2 1.30 670 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). Number before addition of any KOH Number (b) after addition of 25.0 mL of KOH Numbe click to edit (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 HO -P OH Phosphorous acidExplanation / Answer
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
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