12/13/2015 11:00 PM A 71.7/100 12/7/2015 07:46 PM Gradebook Print n Calculator P
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12/13/2015 11:00 PM A 71.7/100 12/7/2015 07:46 PM Gradebook Print n Calculator Periodic Table Question 16 of 17 Map A General Chemistry University Science Books presented by Sapling Learning Donald McQuarrie .Peter A. Rock.Ethan Gallogly Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is pKa1 pKa2 an important compound in industry and agriculture. 1.30 670 Calculate the pH for each of the following points in the titration of 50.0 mL of a 3.00 M H3PO3(aq) with 3.0 M KOH(aq). Phosphorous acid Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH l 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 O Previous Give Up & View Solution Check Answer Next Exit HintExplanation / Answer
H3PO3 millimoles = 50 x 3 = 150
(a) Before any addition of KOH :
H3PO3 ---------------------> H+ + H2PO3-
3 0 0
3 - x x x
Ka1 = [H+][H2PO3-] / [H3PO3]
0.05 = x^2 / 3 - x
x = 0.363
[H+] = 0.363 M
pH = -log [H+] = -log [0.363]
= 0.44
pH = 0.44
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 = 50 x 3 = 150
HPO3^-2 molarity = 150 / (50 +100) = 1
HPO3^-2 + H2O ------------------> H2PO4- + OH-
1 -x x x
Kb2 = x^2 / 1.4-x
5.01 x 10^-8 = x^2 / 1-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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