The Kb for SH- is the equilibrium constant for which of the following reactions? A. SH-+OH-S2+H2O B. SH-+H2OH2S+OH- C. SH-+H3O H2S+H2O D. SH-+H2OH2-+H3O+ Do I need to do a Kw reaction??

Question

The Kb for SH- is the equilibrium constant for which of the following reactions? A. SH-+OH-<>S2+H2O B. SH-+H2O<>H2S+OH- C. SH-+H3O<> H2S+H2O D. SH-+H2O<>H2-+H3O+ Do I need to do a Kw reaction??

.sam. · Answer

I think I got the answer, what's your answer? :)

.sam. · Answer

You don't need Kw, you just need to tinker about it.

anonymous · Answer

can you help me tinker? haha I just need a little guidance.

anonymous · Answer

Do you first put the equations in their equilibrium state?

.sam. · Answer

From the reactant side, you can see that for A, there are 2 negatives, so that won't cancel out.

anonymous · Answer

is the answer D??

.sam. · Answer

Ions are not balanced for D

anonymous · Answer

so then that leaves C

.sam. · Answer

Yes  SH-+H3O+<-> H2S+H2O

Note that hydronium has a positive charge

anonymous · Answer

Thank you. What is the difference if it would be a Ka reaction then?

.sam. · Answer

Ka is for acids, Kb is for bases such as NH3 + H2O <->  NH4+ + OH-

anonymous · Answer

so if the answers to choose from are the same as the ones for this problem then it would be D right? Because the hydronium ion is always a product in acids.

.sam. · Answer

No you still need to consider for the charges of both sides
SH-+H2O<>H2-+H3O+

LHS has 1-
RHS has 0

anonymous · Answer

so do the charges have to sum up to be neutral?

.sam. · Answer

Yes, always.

anonymous · Answer

Would the answer be B??

anonymous · Answer

I got B on this question

anonymous · Answer

B is the only one that I can find neutral that would be a Ka reaction.

anonymous · Answer

This is difficult to do without being able to do an ICE table

anonymous · Answer

Yes I agree.

.sam. · Answer

Ahh yes, the charges cancel, and we're getting base $K_b$
$$\Huge SH^-+H_2O \rightleftharpoons H_2S+OH^-$$
$$K_b=\frac{[H_2S][OH^-]}{[SH^-][H_2O]}$$
Then we can use 
$$K_W=K_aK_b$$

anonymous · Answer

We are looking for Ka now!!! We decided Kb was C.

.sam. · Answer

Yeah use this for conversion
$K_W=K_aK_b$

anonymous · Answer

ok gotcha. thanks. I was confused.

anonymous · Answer

We have no values

.sam. · Answer

Kw=[H+][OH-]

anonymous · Answer

but the answer is b. correct?

anonymous · Answer

does the Ka equal the Kb?

.sam. · Answer

Well we can setup
$$[H^+]\cancel{[OH^-]}=K_a\frac{[H_2S]\cancel{[OH^-]}}{[SH^-][H_2O]}$$
I'm not sure for the rest of the steps