Question

calculate the pH at the equivalence point of a titration of 0.3326 g THAM pKa 5.92 dissolved initially in 25 mL of H2O with 0.1042 M Hcl please show step by step

Answer 1

THAM moles = mass / Molar mass of THAM

            = 0.3326 g / 121.14 g/mol = 0.0027456

at equivalence point acid THAM moles = HCl moles = 0.0027456

Molarity of HCl = moles of HCl / volume of HCl

0.1042 = 0.0027456 / vol

vol of HCl = 0.02635 L

Total solution volume = 0.025 L +0.02635 = 0.05135

at equivalence point all Basic form of THAM (A-) gets converted to protonated form Let it be HA

now we have reverse equilibrium

HA (aq) <--> H+ (aq) + A-(aq) , pka = 5.92 , Ka =10^ -5.92 = 1.2 x 10^ -6

at equilibrium HA moles = 0.0027456-X , H+ = A- moles = X

[HA] = (0.0027456-X) / ( 0.05135 ) , [H+] =[A-] = X /0.05135                    

Ka = [H+] [A-] /[HA]

1.2 x 10^ -6 = (X/0.05135) ( X/0.05135) / ( 0.0027456-X) / ( 0.05135)

1.2x10^-6 = X^2 / ( 0.0027456-X) (0.05135)

X^2+ 6.174 x 10^ -8 -1.69 x 10-10 = 0

X = 1.297 x 10^ -5 = H+

[H+] = ( 1.297 x 10^ -5) / ( 0.05135) = 2.526 x 10^ -4

pH = -log [H+] = -log ( 2.526 x 10^ -4) = 3.6