Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound
ID: 941149 • 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).
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
Explanation / 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
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