For 10 mL of 0.40 M hypophosphorous (structure shown below; pKa = 1.23) give the

Question

For 10 mL of 0.40 M hypophosphorous (structure shown below; pKa = 1.23) give the pH at the various points indicated when titrated with 0.30 M KOH. Kw = 1 x 10-14

Either using the graph paper on the back of this page or using excel. Plot the titration curve with proper title and axis labels for Caffeine showing equivalence point and buffer zone

A. Before any KOH has been added '

B. at 1/6 the equivalence point

C. At ½ the equivalence point

D. At the equivalence point

E. at 5 mL past the equivalence point

Explanation / Answer

pH calculation

A. Before any KOH is added

HA <==> H+ + A-

let x amount has dissociated

Ka = 0.059 = x^2/0.4-x

let x be a small amount then,

x^2 + 0.059x - 0.0236 = 0

x = [H+] = 0.127 M

pH = -log[H+] = 0.90

B. At 1/6th equivalence point

equivalence point

moles of acid present = moles of base added

moles of acid = 0.4 M x 0.01 L = 0.004 mols

1/6th = 0.004/6 = 6.67 x 10^-4 mols

moles of base added = 6.67 x 10^-4 mols

Volume of base added = 6.67 x 10^-4/0.3 = 2.22 x 10^-3 L

molar cnncentration of base = 6.67 x 10^-4/(2.22 x 10^-3 + 0.01) = 0.0546 M

moles of acid remaining = 3.33 x 10^-3 mols

molar concentration of acid = 3.33 x 10^-3/(2.22 x 10^-3 + 0.01) = 0.272 M

pH = pKa + log([base]/[acid])

      = 1.23 + log(0.054/0.272)

      = 0.536

C. At 1/2 equivalence point

moles of acid = moles of salt

pH = pKa = 1.23

D. At equivalence point

moles of salt formed = 0.004 mols

Volume of base added = 0.004/0.3 = 0.0133 L

molar concentration of salt = 0.004/(0.0133 + 0.01) = 0.017 M

A- + H2O <==> HA + OH-

Kb = Kw/Ka = 1.7 x 10^-13 = x^2/0.017

x = [OH-] = 5.37 x 10^-8 M

pOH = -log[OH-] = 7.27

pH = 14 - pOH = 6.73

E. at 5 ml past equivalence point

moles of base = 0.3 x 0.005 = 0.0015 mols

Volume of base added = 0.0183 L

molar concentration of base = 0.0015/(0.0183 + 0.01) = 0.053 M

pOH = -log[OH-] = -log(0.053) = 1.276

pH = 14 - pOH = 12.72

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