I\'m fine with part i) but could someone try part ii) and share their working? T

Question

I'm fine with part i) but could someone try part ii) and share their working? Thanks

A weak diprotic acid, H_2 X, has the following acid dissociation constants. pK_a1 = 2.5 pK_a2 = 7.5 (i) Draw the alpha versus pH speciation diagram for the acid clearly showing the pH values when alpha = 0.5. Label the diagram clearly showing which species corresponds to which line. Explain your reasoning. (ii) 0.0200 moles of H_2X are dissolved in water and the volume made up to 250 mL in a volumetric flask. 10.0 mL aliquots this solution were titrated with a 0.09467 of M solution of NaOH to a thymolphthalein (colour change occurs between pH 9-10.5) endpoint. Predict the endpoint volume of NaOH for the titration explaining your reasoning throughout.

Explanation / Answer

ii) Titration of H2X with NaOH

initial molarity of H2X solution = 0.02 mol/0.25 L

                                                 = 0.08 M

10 ml aliquot of this solution titrated with NaOH

moles H2X present = 0.08 M x 10 ml = 0.8 mmol

moles of NaOH needed for this diprotic acid = 2 x 0.8 = 1.6 mmol

Volume of NaOH needed to reach end point = 1.6 mmol/0.09467 M

                                                                     = 16.90 ml

pH calculation at end point

[Na2X] formed = 0.8 mmol/26.90 ml

                        = 0.03 M

X^2- + H2O <==> HX- + OH-

let x amount has hydrolyzed

Kb1 = Kw/Ka2 = [HX-][OH-]/[X^2-]

1 x 10^-14/3.16 x 10^-8 = x^2/0.03

x = [OH-] = 9.74 x 10^-5 M

pOH = -log[OH-] = 4.01

pH = 14 - pOH = 9.99

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