General Chemistry 4th Edition University Science Books presented by Sapling Leam
ID: 1033156 • Letter: G
Question
General Chemistry 4th Edition University Science Books presented by Sapling Leaming Map osphorous acid, H3PO3(aq), is a diprotic oxyacid that is pKpKa 1.30 6.70 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 Phosphorous acid (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 KOHD Number (e) after addition of 100.0 mL of KOHExplanation / 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 = 0.60
b) after addition of 25.0 mL KOH
it is first half 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 1.5 = 150
HPO3^-2 molarity = 150 / (50 +100) = 1.0M
HPO3^-2 + H2O ------------------> H2PO4- + OH-
1. -x x x
Kb2 = x^2 / 1.4-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 (2.24 x 10^-4 )
pOH = 3.65
pH + pOH = 14
pH = 10.35
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