MnS(s) + 2 H+ --> Mn2+ + H2S(g)

At 25 C the solubility product constant, Ksp, for MnS is 5 x 10-15 and the acid dissociation constants K1 and K2 for H2S are 1 x 10-7 and 1 x 10-13, respectively. What is the equilibrium constant for the reaction represented by the equation above at 25 C?

a) 1x10^-13/5x10^-15
b) 5x10^-15/1x10^-7
c) 1x10^-7/5x10^-20
d)5x10^-15/1x10^-20
e) 1x10^-20/5x10^-15

I know the answer is D but can someone explain why it is this answer?

Keq = (Mn^2+)(H2S)/(H^+)^2

Ksp = (Mn^2+)(S^2-) = 5E-15
H2S ==> 2H^+ + S^2-
k1k2 for H2S = 1E-20

Divide Ksp/k1k2 and see that it produces Keq from above.

(Mn^2+)(S^2-)(H2S)/(H^+)^2(S^2-) and (S^2-) csncels to leave Keq.
Ksp/k1k2 =

cnwtp.amwuj

Well, let's take a closer look at the equation and figure it out together!

The equation represents a precipitation reaction where manganese sulfide (MnS) solid is reacting with two hydrogen ions (2H+) to form manganese ions (Mn2+) and hydrogen sulfide gas (H2S).

Now, in order to determine the equilibrium constant for this reaction, we need to express it in terms of the concentrations of the reactants and products.

The solubility product constant (Ksp) for MnS tells us about the solubility of MnS. It's given as 5 x 10^-15, representing the equilibrium concentration values of Mn2+ and H2S in a saturated solution of MnS at 25°C.

On the other hand, the acid dissociation constants (K1 and K2) for H2S tell us about the degree to which H2S dissociates into H+ and HS- or H+ and S2- ions in water.

Now, by applying the principle of chemical equilibrium, we can determine the equilibrium constant (Keq) for the given reaction.

The equation can be written as:

Keq = [Mn2+] / [H+]^2 x [H2S] / [MnS]

The concentrations of H+ and MnS remain constant, so we can substitute their values directly into the equation. If we do so, we get:

Keq = [Mn2+] / [H2S]

And as we know, the equilibrium concentrations of Mn2+ and H2S are equal to the solubility product constant Ksp for MnS.

So, Keq = Ksp = 5 x 10^-15

Therefore, the answer is option D: 5x10^-15/1x10^-20.

I hope that clarifies things for you! If you have any more questions, feel free to ask!

To determine the equilibrium constant for the given reaction, you need to write the balanced chemical equation and then use the principles of chemical equilibrium.

The balanced equation is:

MnS(s) + 2 H+ → Mn2+ + H2S(g)

According to the law of mass action, the equilibrium constant expression can be written as:

K = [Mn2+][H2S] / [MnS][H+]^2

Now, we need to express the concentrations of the species in terms of the solubility product constant (Ksp) for MnS and the acid dissociation constants (K1 and K2) for H2S.

For MnS(s), since it is a solid, its concentration remains constant, so it can be represented as [MnS] = Ksp.

For H2S, it is a gas and its concentration can be expressed in terms of the partial pressure, which is related to the concentrations of H+ ions and H2S using the acid dissociation constants.

[H2S] = [H+]^2 / K1

Substituting these values into the equilibrium constant expression, we have:

K = [Mn2+][H+]^2 / (Ksp) * ([H+]^2 / K1)

Simplifying further, we get:

K = [Mn2+][H+]^2 / (Ksp * K1)

Now let's substitute the given values:

- Ksp for MnS is 5 x 10^-15
- K1 for H2S is 1 x 10^-7

Substituting these values into the equation, we get:

K = [Mn2+][H+]^2 / (5 x 10^-15 * 1 x 10^-7)

The given reaction doesn't provide information about the concentration of Mn2+. Therefore, we assume it to be 1 (as it doesn't affect the overall equilibrium constant).

Substituting Mn2+ = 1, we have:

K = 1 x [H+]^2 / (5 x 10^-15 * 1 x 10^-7)
K = [H+]^2 / (5 x 10^-22)

And this matches the form D: 5 x 10^-15 / 1 x 10^-20.

So, the equilibrium constant for the reaction is 5 x 10^-15 / 1 x 10^-20.

iuguy