Question

8

Question 8 Tries remaining: 2 Points out of 1.00 P Flag question What is the poH of a 0.05611 M solution of LiHS? The Ką of H2S is 8.9x10-8 Multiple tries are permitted; however, 20% (1/5) point will be deducted for each incorrect response.

Answer 1

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)/8.9*10^-8

Kb = 1.124*10^-7

HS- dissociates as

HS- + H2O -----> H2S + OH-

0.0651 0 0

0.0651-x x x

Kb = [H2S][OH-]/[HS-]

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 ((1.124*10^-7)*6.511*10^-2) = 8.553*10^-5

since c is much greater than x, our assumption is correct

so, x = 8.553*10^-5 M

use:

pOH = -log [OH-]

= -log (8.553*10^-5)

= 4.0679

Answer: 4.07