d/ibis/view.php?id 3822155 PrintCalculator Periodic Table on 17 of 21 eral Chemi
ID: 1033109 • Letter: D
Question
d/ibis/view.php?id 3822155 PrintCalculator Periodic Table on 17 of 21 eral Chemistry 4th Edition University Science Books presented by Sapling Leaming Phosphorous acid, HsPOs(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. PKat pK 1.30 6.70 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 (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH (c) after addition of 50.0 mL of KOH (d) after addition of 75.0 ml of KOH 0 (e) after addition of 100.0 mL of KOH Number Number NumberExplanation / Answer
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 1.5 = 15
HPO3^-2 molarity = 15 / (50 +100) = 0.1 M
HPO3^-2 + H2O ------------------> H2PO4- + OH-
0.1 -x x x
Kb2 = x^2 / 0.1-x
5.01 x 10^-8 = x^2 / 0.1-x
x^2 + 5.01 x 10^-8 x - 5.01 x 10^-9 = 0
x = 7.08 x 10^-5
[OH-] = 7.08 x 10^-5 M
pOH = -log[OH-] = -log (7.08 x 10^-5 )
pOH = 4.15
pH + pOH = 14
pH = 9.85
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.