Question

12. Lactic acid (HC3H5O3) is present in sour milk It is a monoprotic acid. In a 0.100 M solution of lactic acid, the pH is 2.44 at 25°C. Calculate Ka and pKa for lactic acid at that temperature. (10)

Answer 1

Solutions given - acd, 0.100M seletion of lattice pH = 2.44 Lattice acid = HC3 H3 Oz Lavice acid is monoprotic acd so dissociate into OH OH OH H 0 (HC₂H5O3(aq) + H₂O(l) = Cg H50 (99) + H3otaq, We know that - pH = -lope [ Hy ot] 2.44 = -logo [H3 ot] [H3O+] = 10-244 1 [43o+] = 3.6x103m / - C3 450 + 43o+ + H₂O NM NM H C3 H5 O3 Initial 0.100M equilibrium (0.100-22) M Here a= 3.63410 M²

because [H₃O+] = n M At equilibrium so [HC3 HSO3) = (0-100 - 2) = (0.100 - 3,63x103) [HC3H53) = 0.09637 M ] ( C3 H50-] = n [C3H50] = 3.63x10-3 m [H304) = 3.63x10.3 mg we know that - ka= [C3H60] [3 ot [HC3 H5O3 ] ] and pra=-leg, [ka] Ka= (3.63x103) (3063x10-3) 0.09637 1.4 x 10-4 = → Ka ka = 1.4x104 Ka = 1.4 X10-4 ans pka=- log soka =-legio (1.4x104) = -log10 (-4) -logso (10-4)= 3.854 pka = 3.854 ans