University Science Books eral Chemistry 4th Edition presented by Sapling Leaming

Question

University Science Books eral Chemistry 4th Edition presented by Sapling Leaming Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is phont pKa an important compound in industry and agriculture. 1.30 6.70 Calculate the pH for each of the following polnts in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). Number (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 mu of KoH O Previous

Explanation / Answer

Titration of H3PO3 with KOH

(a) Before addition of KOH

H3PO3 + H2O <==> H2PO3- + H3O+

let x amount has dissociated

pKa = -log[Ka]

Ka1 = 5.01 x 10^-2

Ka1 = [H2PO3-][H3O+]/[H3PO3]

5.01 x 10^-2 = x^2/(2.7 - x)

x^2 + 5.01 x 10^-2x - 0.1353 = 0

x = [H3O+] = 0.3436 M

pH = -log[H3O+] = 0.464

(b) after 25 ml of 2.7 M KOH was added

initial moles of H3PO3 = 2.7 M x 50 ml = 135 mmol

moles of KOH added = 2.7 M x 25 ml = 67.5 mmol

moles of H3PO3 left = 67.5 mmol

moles of H2PO3- formed = 67.5 mmol

this is first half-equivalence point

pH = pKa1 = 1.3

(c) after addition of 50 ml of 2.7 M KOH

moles of H3PO3 = moles of KOH added

This is first equivalence point

pH = 1/2(pKa1 + pKa2)

      = 1/2(1.3 + 6.70)

      = 4.00

(d) after 75 ml of 2.7 M KOH added

moles of KOH = 2.7 M x 75 ml = 202.5 mmol

moles of H2PO3- left = 67.5 mmol

moles of HPO3^2- formed = 67.5 mmol

this is second half-equivalence point

pH = pKa2 = 6.70

(e) after 100 ml of 2.7 M KOH added

moles of H3PO3 = 135 mmol

moles of KOH added = 2.7 x 100 = 270 mmol

[HPO3^2-] formed = 135 mmol/(50 + 100) = 0.9 M

HPO3^2- + H2O <==> H2PO3- + OH-

let x amount has hydrolyzed

Kb1 = Kw/Ka2 = [H2PO3-][OH-]/[HPO3^2-]

1 x 10^-14/2 x 10^-7 = x^2/0.9

x = [OH-] = 2.12 x 10^-4 M

pOH = -log[OH-] = 3.673

pH = 14 - pOH = 10.33

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