Question on Equilibrium shift

[05med05]

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Hi,

I encountered a question on the GS3 test which asked the following:

By adding protons to the following reaction, which way would the equilibrium shift?

Ag2S + 4HCN- --> 2{Ag(CN)2}- + S2-

The solution was as follows:

Given that Ka1 and Ka2 for H2S is very small, by adding H2S, you would shift the equilibrium to the right and by adding protons you would shift the equilibrium to the left.

Why? Wouldn't adding more protons lead to a lower concentration of S2- (since more H2S would be formed), thus shifting the equilibrium to the right, in order to create more S2- ?

I'm confused. Any ideas??

 

[willthatsall]

I'm not sure I'm understanding the question that is asked. Where are the hydrogens on the right side of the equation?

 

[Swack]

Hi,

I encountered a question on the GS3 test which asked the following:

By adding protons to the following reaction, which way would the equilibrium shift?

Ag2S + 4HCN- --> 2{Ag(CN)2}- + S2-

The solution was as follows:

Given that Ka1 and Ka2 for H2S is very small, by adding H2S, you would shift the equilibrium to the right and by adding protons you would shift the equilibrium to the left.

Why? Wouldn't adding more protons lead to a lower concentration of S2- (since more H2S would be formed), thus shifting the equilibrium to the right, in order to create more S2- ?

I'm confused. Any ideas??

H2S? Your equation shows HCN 😕

 

[ComicBookHero20]

best way to tackle this question is this

as the equation looks, that acid in the left is relatively strong since it completely dissociates

"Ag2S + 4HCN- --> 2{Ag(CN)2}- + S2-" <-- that is what the question had

but the full rxn, i think looks like:

Ag2S + 4HCN- --> 2{Ag(CN)2}- + S2- + H+

i know it disscociates (near completely, i dunno for sure) is because the Ag(CN)2 forms... this means that the HCN will dissociate into H+ and CN- (the CN- will then bond with Ag)....this is just a double displacemetn reaction

what the question (purposely) left out was that there is H+ on the right side (bc it completely dissocates)

so think le chatlier's

increase H+ (and its on the right), will shift the rxn to the left.

 

[05med05]

But doesn't the S2- combine with H+ to form H2S?

The original question stated that:

Ka1 and Ka2 for H2S (not HCN) is 9.1 x 10 exponent (-8) and 1.2 x 10 exponent (-15) respectively (therefore quite low and H2S would not dissociate).

So then wouldn't it operate the same in reverse? (ie) By adding H+, the S2- would combine with the H+ (to create H2S) leading to a lower concentration of S2- on the products side and then push the equilibrium to the right (so that more S2- forms)?

If what you are saying about the equation is true, and that it actually should look like:

Ag2S + 4HCN- --> 2{Ag(CN)2}- + S2- + H+

Then the part about the H2S ka1 ka2 in the question stem was probably put there to lead one astray... Either that, or the solution provided by GS was questionable, becasue it said that H2S would NOT dissociate, and you are saying that it completely dissociates in solution...

 

[ComicBookHero20]

i think ur thinking about it too in depth, ur bringin up valid points, i agree

the way i think about it is

first this happens: Ag2S + 4HCN- --> 2{Ag(CN)2}- + S2- + H+

then the step: H+ + S2- -> H2S (COULD happen but u dont need to think about it, like its extraneous, bc the Ag(CN)2 forming is the more important one)