Question

We have titrated natural water in which the primary buffering agent is carbonate species and understood how to predict the titration curve using the equilibrium constants of those weak acids. Plot the estimated curve for the titration of 1 L of 0.001 M phosphoric acid with 0.1 M NaOH.

Answer 1

Titration

1 ml NaOH added

[H3PO4] remained = 0.001 M x 1 L - 0.1 M x 0.001 L = 0.0009 mol

[H2PO4-] formed = 0.1 M x 0.001 L = 0.0001 mol

pH = pKa1 + log(base/acid)

     = 2.15 + log(0.0001/0.0009) = 1.20

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2.5 ml NaOH added

[H3PO4] remained = 0.001 M x 1 L - 0.1 M x 0.0025 L = 0.00075 mol

[H2PO4-] formed = 0.1 M x 0.0025 L = 0.00025 mol

pH = pKa1 + log(base/acid)

     = 2.15 + log(0.00025/0.00075) = 1.70

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5 ml NaOH added

[H3PO4] remained = 0.001 M x 1 L - 0.1 M x 0.005 L = 0.0005 mol

[H2PO4-] formed = 0.1 M x 0.005 L = 0.0005 mol

half-equivalence point

pH = pKa1 = 2.15

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7.5 ml NaOH added

[H3PO4] remained = 0.001 M x 1 L - 0.1 M x 0.0075 L = 0.00025 mol

[H2PO4-] formed = 0.1 M x 0.0075 L = 0.00075 mol

pH = pKa1 + log(base/acid)

     = 2.15 + log(0.00075/0.00025) = 2.63

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10 ml NaOH added

[H2PO4-] formed = 0.1 M x 0.010 L = 0.0010 mol

First equivalence point

pH = 1/2(pKa1 + pKa2) = 1/2(2.15 + 7.20) = 4.7

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15 ml NaOH added

[H2PO4-] remained = 0.1 M x 0.015 L - 0.001 M x 1 L = 0.0005 mol

[HPO4^2-] formed = 0.1 M x 0.005 L = 0.0005 mol

Second half-equivalence point

pH = pKa2 = 7.20

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20 ml NaOH added

[H2PO4] remained = 0.1 M x 0.020 L - 0.001 M x 1 L = 0.001 mol

[HPO4^2-] formed = 0.1 M x 0.010 L = 0.001 mol

second equivalence point

pH = 1/2(pKa2 + pKa3) = 1/2(7.20 + 12.35) = 9.8

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25 ml NaOH added

Third half-equivalence point

pH = pKa3 = 12.35

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30 ml NaOH added

Third equivalence point

[PO4^3-] formed = 0.001 mol/1.030 L = 0.00097 M

Kb1 = 0.022 = x^2/0.00097

x = [OH-] = 4.64 x 10^-3 M

pOH = -log[OH-] = 2.33

pH = 14 - pOH = 12.00

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Titration plot below