Question

the pH of .040 M weak base B is 10.78. what is value of Kb B?

Answer 1

Ionization Equation for base B :-

B + H₂O --------------> HB⁺ + OH^-

PAGEI Giveno- Concentration of weak base B = 0.04M. pH of of lease = 10.78. To calculate : - Kb for bоце ? calculation :- Ionization equation for bese Bi- B + H2O HBT tot Kb = [HBt] [oH] 18] - Calculation of teet) {H} :- pH = -log [ 4t ] or St = 10 PH [H+] = 10-10 [ut] = 10659 x10 "M Kw = [H] xLot) Kw = 1.00 x 1014 (Universal value) 1.0081014 = 1.659 X10 X804] {OH] = 1.00 x 10" = 10 1.659 x 1011 [0] = 0.602x10²m] - Calculation of (HB*}:- from the ionization equation, we know there is 1:1 molar ratio between {HBt] and [OHJ. Therefore, {HB+] = [OH-] = 0.602x103M. Now,

PAGE 2 - The value in the of [B] is (B] = given 0.04 M. question le. Son Kb = {HBt] [H] [B] Kb = 0.60 2x10 3 x 0.6024103 0.04 kb = 0.3624 x 106 - 0.04 I kb= 9.06810 6 And The value of kb for weak base B as 9.068106