Question

What is the pOH of a 0.01743 M solution of LiHS? The Ka of H2S is...

What is the pOH of a 0.01743 M solution of LiHS? The Ka of H2S is 8.9×10^–8.

0 0
Add a comment Improve this question Transcribed image text
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-

0.0174 0 0

0.0174-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)*1.743*10^-2) = 4.425*10^-5

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

so, x = 4.425*10^-5 M

use:

pOH = -log [OH-]

= -log (4.425*10^-5)

= 4.354

Answer: 4.35

Add a comment
Know the answer?
Add Answer to:
What is the pOH of a 0.01743 M solution of LiHS? The Ka of H2S is...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT