Question 4 of 5 University Science Books General Chemistry 4th Editio Donald McQ

Question

Question 4 of 5 University Science Books General Chemistry 4th Editio Donald McQuarrie Peter A. Rock. Ethan Gallogly presented by Sapling Learning Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is pKa pKa2 an important compound in industry and agriculture. 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 Number (c) after addition of 50.0 mL of KOH l Number (d) after addition of 75.0 mL of KOH l Number e) after addition of 100.0 mL of KOH HO P OH Phosphorous acid Map

Explanation / Answer

pKa = -log[Ka]

Titration

(a) Before addition of KOH

H3PO3 <==> H2PO3- + H+

let x amount has dissociated

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

0.050 = x^2/(1.5 - x)

x^2 + 0.050x - 0.075 = 0

x = [H+] = 0.25 M

pH = -log[H+] = 0.60

(b) after 25 ml of KOH added

moles of H3PO3 = moles of H2PO3-

This is first half equivalence point

pH = pKa1 + log(base/acid)

      = 1.30 + log([H2PO3-]/[H3PO3])

      = 1.30

(c) after 50 ml of KOH added

moles of H3PO3 = moles of KOH added

This is first equivalence point

pH = 1/2(pKa1 + pKa2) = 1/2(1.30 + 6.70) = 4.00

(d) after 75 ml of KOH added

moles of H2PO3- = moles of HPO3^2-

This is second half equivalence point

pH = pKa2 = 6.70

(e) after 100 ml of KOH added

moles of KOH is twice the moles of H3PO3 present

this is second equivalence point

formed [HPO3^2-] = 1.5 M x 50 ml/150 ml = 0.5 M

salt hydrolyzes,

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

let x amount has reacted

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

1 x 10^-14/2 x 10^-7 = x^2/0.5

x = [OH-] = 1.58 x 10^-4 M

pOH = -log[OH-] = 3.80

pH = 14 - pOH = 10.2

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