Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound
ID: 918079 • Letter: P
Question
Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture.
Phosphorous add, H_3P0_3(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 2.7 M H_3PO_3(aq) with 2.7 M KOH(aq). before addition of any KOH after addition of 25.0 mL of KOH after addition of 50.0 mL of KOH after addition of 75.0 mL of KOH after addition of 100.0 mL of KOHExplanation / Answer
a) before addition of KOH :
H3PO3 -----------------> H+ + H2PO3-
2.7 0 0
2.7 - x x x
Ka1 = [H+][H2PO3-] / [H2PO3]
0.05 = x^2 / 2.7 - x
x = 0.343
[H+] = 0.343 M
pH = -log[H+] = -log(0.343)
pH = 0.46
1) after addition of 25.0 mL KOH
it is first equivaelce point
pH = pKa1 = 1.30
pH = 1.30
2 ) 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
3) 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 = 50 x 2.7 = 135
HPO3^-2 molarity = 135 / (50 +100) = 0.9 M
HPO3^-2 + H2O ------------------> H2PO4- + OH-
0.9 -x x x
Kb2 = x^2 / 0.9 -x
5.01 x 10^-8 = x^2 / 0.9-x
x^2 + 5.01 x 10^-8 - 7.017 x 10^-8 = 0
x = 2.12 x 10^-4
[OH-] = 2.12 x 10^-4 M
pOH = -log[OH-] = -log (2.12 x 10^-4 )
pOH = 3.67
pH + pOH = 14
pH = 10.33