Question

Citric Acid is a tri-protic acid with Ka1 = 8.4 x 10^-4 and Ka2 = 1.8 x 10^-5 and Ka3 = 4.0 x 10^-6. a) Calculate the pH at the 2nd equivalence point in the titration of 100 mL of a 0.21 M Citric Acid solution with 0.25 M NaOH. b) Calculate the pH, after 15.8 mL of NaOH have been added. c) Calculate the pH, 5 mL after the LAST equivalence point.

Answer 1

Answering (a) and (b)

H₃Cit + H₂O    H₂Cit^- + H₃O⁺

H₂Cit^-+ H₂O     HCit^2- + H₃O⁺

HCit^2- + H₂O     Cit^3- + H₃O⁺

Addition of NaOH will first convert, H₃Cit in to H₂Cit^- until all the H₃cit is used up then H₂cit^– will be converted into Hcit₂^–, etc.)

Given Ka1 = 8.4 x10^-4 => pKa1 = - logKa1 = - log(8.4x10^-4) = 3.076

Ka2 = 1.8x10^-5 => pKa2 = - logKa2 = - log(1.8x10^-5) = 4.745

Ka3 = 4.0x10^-6 => pKa3 = - logKa3 = - log(4.0x10^-6) = 5.4

(a) pH at the second equivalence point = (pKa2 + pKa3) / 2 = 4.745 + 5.4 /2 = 5.07

(b) Initial Moles of Citric Acid = (100 /1000) Lt * 0.21 M = 0.021 moles

Moles of NaOH added (In 15.8 mL volume) = (15.8 /1000) Lt * 0.25 M = 0.00395 moles

Since addition of NaOH will first convert, H₃Cit in to H₂Cit^-

0.00395 moles of NaOH will consume 0.00395 moles of H₃Cit and generate 0.00395 moles of H₂Cit^{- .}

Now , [H₃Cit ] = 0.021 moles - 0.00395 moles = 0.01705 moles

[H₂Cit^-] = 0.00395 moles

H₃Cit + H₂O    H₂Cit^- + H₃O⁺ (After adding 15.8 ml of NaOH added)

0.021 moles 0 0 0

0.01705 moles   0.00395 moles   

pH = pKa + log [H₂Cit^-] /  [H₃Cit ]

= 3.076  + log (0.00395) / (0.01705)

= 3.076 + log (0.231) = 3.076 + (- 0.63)

pH = 2.446