Question

To obtain the data needed to plot a titration curve for the titration of a weak acid with a strong base, a chemist used 50.00 mL of 0.10mol/L of citric acid with 0.10 mol/L NaOH(aq). Calculate the pH of the resulting solution at the equivalent point.

Answer 1

The citric acid is a weak organic tribasic acid. It's chemical formulae can be written for simplicity as AH₃. Therefore, the reaction of the titration of the citric acid by sodium hydroxide is:

AH₃ + 3NaOH $\rightarrow$ Na₃A + 3H₂O.

According to the stoichiometric coefficients, the number of the moles of NaOH added and the number of the moles of citric acid in the sample relate as:

$\frac{n(NaOH)}{3} = n(AH_3)$ .

Therefore, at the equivalence point, the volume of NaOH added is:

$V = \frac{n}{c} = \frac{3c(AH_3)V(AH_3)}{c(NaOH)} = \frac{3·0.10·50.00}{0.10} = 150.00$ mL.

The concentration of sodium citrate at the equivalence point is:

$c(Na_3A) = \frac{c(AH_3)V(AH_3)}{V_{tot}} = \frac{0.10·50.00}{150.00+50.00} = 0.025$ mol/L.

The citrate anion partly reacts with water:

A^3- + H₂O $\rightarrow$ HA^2- + OH^-.

According to this reaction, OH^- anions are released. The equilibrium constant of this reaction, $K_b$ , relates to the third step acidity constant of citric acid as:

$K_b = \frac{K_w}{K_{a3}} = 2.5·10^{-8}$ .

As there is two orders difference between $K_{a2}$ and $K_{a3}$, we can consider only this reaction (the third step of dissociation).

The concentration of OH^- anions is therefore:

$K_b = \frac{[HA^{2-}][OH^-]}{[A^{3-}]}$ .

The equilibrium concentration of OH^- anions equals the equilibrium concentration of HA^2- anions. The equilibrium concentration of the citrate anion is (0.025 - x), where x is the equilibrium concentration of OH^-:

$(0.025-x)·2.5·10^{-8} = x^2$

$x = 2.5·10^{-5}$

$[OH^-] = 2.5·10^{-5}$

$pOH = 4.6$

$pH = 14-pOH = 9.4$

Answer: the pH of the resulting solution at the equivalence point is 9.4.