Question

Calculate the pH at the second stoichiometric point when 130 mL of a 0.030 M solution of glutaric acid (K_a1=4.54*10^-5, K_a2=5.4*10^-6) is titrated with 1.0 M NaOH.

Answer 1

No of mole of Glutoric acid =( 0.03mol/1000ml)× 130ml =0.0039

No of mole of NaOH required = 0.0039 ×2 =0.0078

Volume of NaOH required = (1000ml/1.0mol)×0.0078 =7.8ml

So, second equivalence point is 7.8ml

at second equivalence point all the Glutomic is converted to Disodium Glutamate

no of mole of Na2Glu = 0.0039

Total volume = 137.8ml

[ Glu^2- ] = (0.0039mol/137.8ml)×1000ml = 0.0283M

Glu^2- is hydrolysed by water

Glu^2-   + H2O ------> HGlu^-   + OH-

Kb = [OH- ] [ HGlu^-]/[Glu^2-]

Kb= Kw/Ka2

= 1×10^-14/5.4×10^-6

= 1.85×10^-9

at equillibrum

[ Glu^2- ] = 0.0283 - x

[ OH- ] = x

[ HGlu- ] = x

x^2/(0.0283 -x) = 1.85 × 10^-9

we can assume

0.0283 - x = 0.0283

x^2/0.0283 = 1.85 ×10^-9

x^2 = 5.23×10^-11

x = 7.23×10^-6

[ OH- ] = 7.23 × 10^-6M

pOH = 5.14

pH = 14 - pOH

= 14 - 5.14

= 8.86