Question

Which of the following statements is true regarding the dissolution of a salt in a solvent? Medal and Fan
A. An equilibrium constant can be described that is the product of the concentrations of the ions to the power of their coefficient
B. The reaction does not have an equilibrium constant since all dissolution reactions are nonreversible
C. Precipiates are included in the equilibrium constant
D. The number of each ion does not affect the equilibrium constant.

Would it be D?

Answer 1

Oh god. I'm terrible at solutions and stuff but yeah, I think you're right.

Answer 2

Yeah, so am I ahah

Answer 3

Because I mean. Ugh. Let's say that we have NaCl, common table salt. It would be 1NaCl, one mole of NaCl contains 1 mole of Na and 1 mole of Cl. So there's the same amount of stuff just in a different fashion. B would also make sense for whatever reason.

Answer 4

I mean can Na and Cl combine to reform NaCl or nah?

Answer 5

The concentrations aren't changing, it's just the ions, which would just change the isotope....so I don't see how that would affect Kc

Answer 6

Hmm. Then yeah, I'd go with D.

Answer 7

Ugh. I'm so nervous.

Answer 8

Oh, you'll do fine Ms. 97 Percent Is My Worst Grade.

Answer 9

93. I said 93 last tri. Lol

Answer 10

Wait maybe it does affect the constant since there are more of the Cl or Na and since there are more reactants, the constant shifts towards the products....

Answer 11

@seehearcreate equlibrium constant only depends upon temperature...

Answer 12

Pressure too. But only with gasses.

Answer 13

And concentration as well.

Answer 14

So, I found this?
NOTE: Unlike Ka and Kb for acids and bases, the relative values of Ksp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions.

Would that mean that D is still correct?

Answer 15

Nevermind, that actually proves D is correct. I think it's C since you multiply by the products and the gas/liquid/ not pure solids law doesn't apply to Ksp

Answer 16

'The number of each ion' affects the concentration and thus (by definition) affect the equilibrium constant. So it's not D.

A. An equilibrium constant can be described that is the product of the concentrations of the ions to the power of their coefficient

Is correct, the solubility product constant \(K_{sp}\) is what is used for the dissociation for salts.

For: \(\large XY_{(s)} \rightleftharpoons a~ X^-_{(aq)}+b~Y^+_{(aq)} \)

It is: \(\large K_{sp}=\dfrac{[X^-]^a[Y^+]^b}{[XY]}\)

but since [XY] is a solid, it is equal to 1.

so the equation can be simply written as \(\large K_{sp}=[X^-]^a[Y^+]^b\)

Answer 17

That's what I actually picked(: Good, I'm glad!!!

Answer 18

@seehearcreate nope eqlb cons only depends on temperature....

Answer 19

No...that isn't true. In an equation that includes gibbs free energy, maybe, but increasing the reactants causes a shift in the equilibrium constant towards the products.

Answer 20

this has no effect on the value of K itself. It does affect the equilibrium, i.e. it shifts it either to the left or the right to compensate for the change until a new equilibrium is achieved. K still remains the same. What really changes is the value of Q, the reaction quotient,

Answer 21

@rishavraj you're right, it doesn't affect the constant value, hence it's a constant.

what i said earlier is incorrect, "The number of each ion' affects the concentration and thus (by definition) affect the equilibrium quotient." fixed.

In hindsight, I think D is \(\sf more\) correct, because the Ksp is called the solubility product constant, not the equilibrium constant explicitly -

\(\sf although, ~it ~is~ the~ exact~ same ~thing,~ just ~a~ different~ name\).