20)
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/1.8*10^-5
Kb = 5.556*10^-10
C2H3O2- dissociates as
C2H3O2- + H2O -----> HC2H3O2 + OH-
0.1 0 0
0.1-x x x
Kb = [HC2H3O2][OH-]/[C2H3O2-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.556*10^-10)*0.1) = 7.454*10^-6
since c is much greater than x, our assumption is correct
so, x = 7.454*10^-6 M
use:
pOH = -log [OH-]
= -log (7.454*10^-6)
= 5.1276
use:
PH = 14 - pOH
= 14 - 5.1276
= 8.8724
Answer: D
20) Determine the pH of a 0.10 M NaC2H302. For acetic acid, Ka = 1.8 x 10-5. A) 4.18 B) 5.29 C) 7.33 D) 8.87
please show work
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