Question

Citric acid is a tri-protic acid with Ka1= 8.4x10^-4, Ka2= 1.8x 10^-5 Ka3= 4.0x 10^-6 a) calculate the pH at the 2dn equlivance point in the titration of 100.0 ml of 0.21M citric acid solution with 0.25M NaOH. Calculate the pH after 15.8 ml of NaOH has been added. c) calculate the pH 5.0 ml after the last equivalence point

Answer 1

Citric acid, H₃A is a triprotic acid. The acid ionization constants are given as

H₃A (aq) -------> H⁺ (aq) + H₂A^- (aq); K_a1 = 8.4*10^-4

H₂A^- (aq) -------> H⁺ (aq) + HA^2- (aq); K_a2 = 1.8*10^-5

HA^2- (aq) --------> H⁺ (aq) + A^3- (aq); K_a3 = 4.0*10^-6

a) At the 2^nd equivalence point, H₃A is completely converted to HA^2- and the system contains HA^2-/A^3- as the acid-base pair. The step-wise neutralization reactions are given as

H₃A (aq) + NaOH (aq) -------> NaH₂A (aq) + H₂O (l)

NaH₂A (aq) + NaOH (aq) --------> Na₂HA (aq) + H₂O (l)

At the 2^nd equivalence point,

1 mole H₃A = 2 moles NaOH.

Millimoles of HA^2- formed = millimoles of H₃A taken = (100 mL)*(0.21 M) = 21.0 mmole.

Millimoles of NaOH required to reach the 2^nd equivalence point = 2*21.0 mmole = 42.0 mmole.

Volume of 0.25 M NaOH required = (42.0 mmole)/(0.25 M) = 168.0 mL.

Total volume of the solution at the 2^nd end point = (100.0 + 168.0) mL = 268.0 mL.

Concentration of HA^2- formed = (millimoles of HA^2-)/(total volume of solution) = (21.0 mmole)/(268.0 mL) = 0.078 M.

HA^2- establishes equilibrium as shown by the last expression above and hence, we can write,

K_a = [H⁺][A^3-]/[HA^2-]

====> 4.0*10^-6 = (x).(x)/(0.078 – x) [The concentrations of H⁺ and A^3- are equal due to the 1:1 nature of ionization.]

====> 4.0*10^-6 = x²/(0.078 – x)

Since K_a is small and [HA^2-] is small, we can assume x << 0.078 M and write

4.0*10^-6 = x²/(0.078)

====> x² = 3.12*10^-7

====> x = 5.5857*10^-7

Thus, [H⁺] = 5.5857*10^-7 and pH = -log [H⁺] = -log (5.5857*10^-7) = 3.2529 ≈ 3.25 (ans).

b) Millimoles of NaOH added = (15.8 mL)*(0.25 M) = 3.95 mmole.

Millimoles of H₃A neutralized = millimoles of H₂A^- formed = millimoles of NaOH added = 3.95 mmole.

Millimoles of H₃A retained = (21.0 – 3.95) mmole = 17.05 mmole.

The volume of the solution stays the same; hence, we can write

[H₂A^-]/[H₃A] = (millimoles of H₂A^-)/(millimoles of H₃A) = (3.95 mmole)/(17.05 mmole)

Use the Henderson-Hasslebahc equation.

pH = pK_a1 + log [H₂A^-]/[H₃A]

====> pH = -log (K_a1) + log (3.95 mmole)/(17.05 mmole)

====> pH = -log (8.4*10^-4) + log (0.23167)

====> pH = 3.07572 + (-0.63513) = 2.44059 ≈ 2.44 (ans).

c) Consider the complete neutralization reaction as below.

H₃A (aq) + 3 NaOH (aq) -------> Na₃A (aq) + 3 H₂O (l)

As per the stoichiometric equation,

1 mole H₃A = 3 moles NaOH.

Millimoles of NaOH required to reach the 3^rd equivalence point = 3*21.0 mmole = 63.0 mmole.

Volume of NaOH required = (63.0 mmole)/(0.25 M) = 252.0 mL.

5.0 mL of excess NaOH is added; hence, the final volume of the solution is (100.0 + 252.0 + 5.0) mL = 357.0 mL.

The solution now contains A^3- and excess NaOH. NaOH is a much stronger base and hence, the pH of the solution is decided by NaOH.

[OH^-] = (5.0 mL)*(0.25 M)/(357.0 mL) = 0.0035 M.

pOH = -log [OH^-] = -log (0.0035) = 2.45593; therefore, pH = 14 – pOH = 14 – 2.45593 = 11.54407 ≈ 11.54 (ans).