In: Chemistry
1. The Ka of a monoprotic weak acid is 6.11 × 10-3. What is the percent ionization of a 0.151 M solution of this acid? use quadratic equation to solve
2. A certain weak base has a Kb of 7.60 × 10-7. What concentration of this base will produce a pH of 10.23?
please show all your steps
thank you!
HA ------> H+ + A-
initial c 0 0
change -c c c
at equilibrium c(1-) c c
is a ionisation constant
Ka = [H+][A-]/[HA]
Ka = c * c/c(1-)
Ka = C2/1- 1<<<<<
Ka = C2
6.11*10-3 = 0.1512
2 = 6.11*10-3 /0.151
= 4.046*10-2
= 0.2011
% = 20.11%
2. PH = 10.23
POH = 14-PH
= 14-10.23 = 3.77
POH = -log[OH-]
3.77 = -log[OH-]
[OH-] = 10-3.77 = 1.7*10-4 M
[OH-] = Kb*c
1.7*10-4 = 7.6*10-7 *C
(1.7*10-4)2 = 7.6*10-7 *C
C = 0.038M
concentraction is 0.038M >>>> answer