# Solubility and equilibrium question

#### sbuxaddict

7+ Year Member
When you are calculating the equilibrium constant, you take the coefficient and apply it as an exponent.

4H2 becomes [H]^4 in the equilibrium expression.

My question is, if Ksp is supposed to be the equilibrium constant for salt dissociating into ions, why is it that when you're calculating solubility product (Ksp), the coefficient is also the coefficient AND exponent in the expression?

ie: Ca3(PO4)2 --> 2Ca + 2PO4
Ksp = [Ca]^3 * [PO4]^3 = 3x^3 * 2x^2

#### ingramw1202

When you are calculating the equilibrium constant, you take the coefficient and apply it as an exponent.

4H2 becomes [H]^4 in the equilibrium expression.

My question is, if Ksp is supposed to be the equilibrium constant for salt dissociating into ions, why is it that when you're calculating solubility product (Ksp), the coefficient is also the coefficient AND exponent in the expression?

ie: Ca3(PO4)2 --> 2Ca + 2PO4
Ksp = [Ca]^3 * [PO4]^3 = 3x^3 * 2x^2
Ca3(PO4)2-----> 3Ca^(2+) + 2(PO4)^(3-)

So: Ksp=[P]/[R] but since reactant is a solid, it does not appear. Therefore,

Ksp= [Ca2+]^3 x [(PO4)^3-]^2

The type of questions you are talking about are those in which you are given a concentration in terms of the molar solubility of the reactant. So for example, the question might say that The molar solubility of Ca3(PO4)2= X They want the answer of Ksp in terms of X. By looking at the reaction, we can see that for every one mole of Ca3(PO4)2 that dissolves, we get 3 moles of Ca and 2 moles of PO4. So This means that the concentration [Ca] can be rewritten as 3X and the concentration [PO4] can be rewritten as 2X. This means Ksp becomes:

Ksp= [3X]^3 x [2x)]^2
Note: The stoichiometric coefficient is NOT the coefficient and the exponent. It only appears this way. What is actually being asked is that you write the Ksp in terms of X when X=[Ca3(PO4)2]. All you really do is write the equation: 1 mol reactant--> 3 mol Ca and 1 mol reactant--> 2 mol of PO4. Well, we know that 1 mol reactant = X. This means that [Ca] = 3X and [PO4]=2X

OP
S

#### sbuxaddict

7+ Year Member
Ca3(PO4)2-----> 3Ca^(2+) + 2(PO4)^(3-)

So: Ksp=[P]/[R] but since reactant is a solid, it does not appear. Therefore,

Ksp= [Ca2+]^3 x [(PO4)^3-]^2

The type of questions you are talking about are those in which you are given a concentration in terms of the molar solubility of the reactant. So for example, the question might say that The molar solubility of Ca3(PO4)2= X They want the answer of Ksp in terms of X. By looking at the reaction, we can see that for every one mole of Ca3(PO4)2 that dissolves, we get 3 moles of Ca and 2 moles of PO4. So This means that the concentration [Ca] can be rewritten as 3X and the concentration [PO4] can be rewritten as 2X. This means Ksp becomes:

Ksp= [3X]^3 x [2x)]^2
Note: The stoichiometric coefficient is NOT the coefficient and the exponent. It only appears this way. What is actually being asked is that you write the Ksp in terms of X when X=[Ca3(PO4)2]. All you really do is write the equation: 1 mol reactant--> 3 mol Ca and 1 mol reactant--> 2 mol of PO4. Well, we know that 1 mol reactant = X. This means that [Ca] = 3X and [PO4]=2X
So what I'm getting from your explanation is that when molar solubility is involved, you write Ksp in terms of X (or did I misunderstand that?)

There's this question in the BR GCHEM book that really threw me off:

What is the molar solubility of CaCl2(s) in 0.01M NaCl(aq) solution?
CaCl2(s) --> Ca2+(aq) to 2Cl-(aq) Ksp = 2.5 * 10^(-10) M^3

Then there's this chart to calculate shifts, etc.

The final equilibrium setup they have:

@ equilibrium: Ca2+ = x, and 2Cl- = 0.01 + 2x

with Ksp = [Ca2+][Cl-]^2 = (x)(0.01+2x)^2

In this instant, why is it not Ksp = [Ca2+]2[Cl-]^2 = (x)2(0.01+2x)^2, since molar solubility is involved?

#### ingramw1202

So what I'm getting from your explanation is that when molar solubility is involved, you write Ksp in terms of X (or did I misunderstand that?)

There's this question in the BR GCHEM book that really threw me off:

What is the molar solubility of CaCl2(s) in 0.01M NaCl(aq) solution?
CaCl2(s) --> Ca2+(aq) to 2Cl-(aq) Ksp = 2.5 * 10^(-10) M^3

Then there's this chart to calculate shifts, etc.

The final equilibrium setup they have:

@ equilibrium: Ca2+ = x, and 2Cl- = 0.01 + 2x

with Ksp = [Ca2+][Cl-]^2 = (x)(0.01+2x)^2

In this instant, why is it not Ksp = [Ca2+]2[Cl-]^2 = (x)2(0.01+2x)^2, since molar solubility is involved?
Let us look at the Ksp given and compare it to the previous question:

Ksp=[Ca2+][Cl-]^2 = (x)(0.01+2x)^2
So as this reaction goes towards equilibrium, the 1 mol of CaCl2 dissociates into 1 mol Ca and 2 mol Cl-. This is what the Ksp in terms of the reaction says. But when you write the concentrations as you showed (x)(0.01+2x)^2, X is the change in concentration of products and reactants. As the reaction proceeds towards equilibrium, the concentration of CaCl2 decreases by X and Ca2+ increases by X and Cl- increases by 2X. But CaCl2 is not dissolving in a solution that contains no products. Instead is dissolved in a solution that already contains 0.01 M Cl. So Cl concentration increases by 0.1 + 2x. Not 2(0.1 +2x).

Previous question: Ksp= [Ca2+]^3 x [(PO4)^3-]^2 for molar solubility of Ca3(PO4)2= X
In the previous question, the solubility of Ca3(PO4)2 is a known number which has a value of X. Ca3(PO4)2 concentration decreases by X, and Ca+ concentration increases by 3X and PO42 concentration increase by 2X. You can sub these into the equilibrium equation and get that at equilibrium, Ksp=[P]/[R] = equilibrium concentrations of products and reactants all raised to stoichiometric coefficients = [3x]^3 x [2x]^2.

In CaCl2 dissociation in 0.01 mol NaCl, at equilibrium Ksp= equilibrium concentrations of products and reactants all raised to stoichiometric coefficients. So at equilibrium in this reaction, CaCl2 concentration decreases by X, Ca concentration increases by X, and Cl concentration increases by 0.01 + 2x, where 0.01 is the concentration of Cl already in the solution. Plugging these values into Ksp=equilibrium products concentration over equilibrium reactant concentration all raised to stoichiometric coefficients gives Ksp= [X] [0.01 + 2x]^2.

So i didn't just put the coefficients in there because its molar solubility. The equilibrium expression is an equation of the reaction at equilibrium. You have to define the concentration of products and reactants in terms of X and sub them into that equation. At equilibrium in question 1, the concentration of each is 3x and 2x respectively, because in a 1 molar solution of Ca3(PO4)2, you have 3 moles of Ca and 2 moles of PO4. In the second question, a 1 molar solution of CaCl2 in a 0.01 molar solution of NaCl contains 1 mole of Ca and (0.01 moles plus 2 moles) of Cl.

Does this make sense?

OP
S

#### sbuxaddict

7+ Year Member
Let us look at the Ksp given and compare it to the previous question:

Ksp=[Ca2+][Cl-]^2 = (x)(0.01+2x)^2
So as this reaction goes towards equilibrium, the 1 mol of CaCl2 dissociates into 1 mol Ca and 2 mol Cl-. This is what the Ksp in terms of the reaction says. But when you write the concentrations as you showed (x)(0.01+2x)^2, X is the change in concentration of products and reactants. As the reaction proceeds towards equilibrium, the concentration of CaCl2 decreases by X and Ca2+ increases by X and Cl- increases by 2X. But CaCl2 is not dissolving in a solution that contains no products. Instead is dissolved in a solution that already contains 0.01 M Cl. So Cl concentration increases by 0.1 + 2x. Not 2(0.1 +2x).

Previous question: Ksp= [Ca2+]^3 x [(PO4)^3-]^2 for molar solubility of Ca3(PO4)2= X
In the previous question, the solubility of Ca3(PO4)2 is a known number which has a value of X. Ca3(PO4)2 concentration decreases by X, and Ca+ concentration increases by 3X and PO42 concentration increase by 2X. You can sub these into the equilibrium equation and get that at equilibrium, Ksp=[P]/[R] = equilibrium concentrations of products and reactants all raised to stoichiometric coefficients = [3x]^3 x [2x]^2.

In CaCl2 dissociation in 0.01 mol NaCl, at equilibrium Ksp= equilibrium concentrations of products and reactants all raised to stoichiometric coefficients. So at equilibrium in this reaction, CaCl2 concentration decreases by X, Ca concentration increases by X, and Cl concentration increases by 0.01 + 2x, where 0.01 is the concentration of Cl already in the solution. Plugging these values into Ksp=equilibrium products concentration over equilibrium reactant concentration all raised to stoichiometric coefficients gives Ksp= [X] [0.01 + 2x]^2.

So i didn't just put the coefficients in there because its molar solubility. The equilibrium expression is an equation of the reaction at equilibrium. You have to define the concentration of products and reactants in terms of X and sub them into that equation. At equilibrium in question 1, the concentration of each is 3x and 2x respectively, because in a 1 molar solution of Ca3(PO4)2, you have 3 moles of Ca and 2 moles of PO4. In the second question, a 1 molar solution of CaCl2 in a 0.01 molar solution of NaCl contains 1 mole of Ca and (0.01 moles plus 2 moles) of Cl.

Does this make sense?
Ohhh gotcha, yes that does.
Thank you so much for writing out such a detailed explanation!