University Science Books General Chemistry 4th Edition presented by Sapling Lear

Question

University Science Books General Chemistry 4th Edition presented by Sapling Learning Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is pKa1 pKa2 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 1.8 M H3PO3(aq) with 1.8 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 100.0 mL of KOH HO P OH Phosphorous acid Map

Explanation / Answer

mmol of acid = MV = 50*1.8 = 90 mmol

so

a) before anny addition

Ka = [H+][H2PO4-]/[H3PO4]

10^-1.3 = x*x/(1.8-x)

x = 0.27634 M

pH = -log(x) = -log(0.27634) = 0.5585

B)

mmol of base = MV = 25*1.8 = 45 mmol

there is half neutralization

H3PO4 + OH- -> H2PO4-

mmol of H3PO4 = 90-45 = 45

mmol of H2PO4- = 45

this is abuffer so

pH = pKa1 + log(H2PO4- / H3PO4 )

pH = 1.3 + log(45/45)

pH = 1.3

c)

after 50 mL...

mmol of acid = 90

mmol of base = 90

this is the FIRST equivalence point, meaning it will be the point between pKa1 and pKa2

pH 1st equivalence = 1/2*(pKa1 + pKA2) = 1/2*(1.3+6.7)= 4

pH = 4

d)

after 75 mL, i.e. 25 mL after first neutralization

50 mmol of H2PO4- react with 25 mmol of OH-

25 mmol of HPO4-2 produces so, once again, buffer zone, but 2nd ionization!

so

pH = pKA2 + log(HPO4-2 / H2PO4-)

pH = 6.7 + log(45/45)

pH = 6.7

e)

finally

mmol o fbase = 100*1.8 = 18 mmol

mmol of acid = 90

this is the 2nd equivalence point...

which can be calculated via:

pH = 1/2*(pKa2 + pKa3) = 1/2*(6.70+ 12.35) = 9.525

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