# Ksp vs Keq question (assistance needed)

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#### tvbbnumber41

##### Full Member
7+ Year Member
Hey guys so I ran into a question that stated.....

Which of the following represents the solubility product of Ca(OH)i when it is the byproduct of the reaction between calcium hypochloritc and water?

I wrote out the eq
Ca(OH)2 --> Ca^+2 + (OH^-)2

so I ended up with Ksp = [Ca] [2OH]^2

but apparently it's wrong and its actually Ksp = [Ca] [OH]^2
why is that?

Okay so from what I've inferred from looking at some sample questions... When determining Ksp the concentrations are known or at least represented as a known variable (incidences which include a strong acid/base and complete dissociation). In this case we do not need to represent concentration with a second unknown variable.
In contrast to questions where we are given Ksp and need to find the concentrations, then we must use a unknown variable thus, these would be situations where the [2x]^2 and whatnot are used. These type of questions also include incidences of "sparingly soluble".

I know my explanation is bad, but somewhat correct?

Here is a similar problem and explanation from the AAMC Chemistry Q-pack, and also a slightly more detailed explanation for the same problem. Try to remember you are relating Ksp and molar solubility.

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Okay so from what I've inferred from looking at some sample questions... When determining Ksp the concentrations are known or at least represented as a known variable (incidences which include a strong acid/base and complete dissociation). In this case we do not need to represent concentration with a second unknown variable.
In contrast to questions where we are given Ksp and need to find the concentrations, then we must use a unknown variable thus, these would be situations where the [2x]^2 and whatnot are used. These type of questions also include incidences of "sparingly soluble".

I know my explanation is bad, but somewhat correct?

This sounds unusually complicated.

Here is a similar problem and explanation from the AAMC Chemistry Q-pack, and also a slightly more detailed explanation for the same problem. Try to remember you are relating Ksp and molar solubility.

That's key. The solubility product constant (Ksp) is an equilibrium constant, so its expression is simply Ksp = [products] since the reactants are undissolved solid and liquid water which aren't included in the equation. For the reaction listed, Ksp = [Ca2+][OH-]^2. If you are calculating molar solubility from Ksp, set S = [Ca2+], which from stoichiometry means 2S = [OH-], resulting in Ksp = 4S^3.

Okay so from what I've inferred from looking at some sample questions... When determining Ksp the concentrations are known or at least represented as a known variable (incidences which include a strong acid/base and complete dissociation). In this case we do not need to represent concentration with a second unknown variable.
In contrast to questions where we are given Ksp and need to find the concentrations, then we must use a unknown variable thus, these would be situations where the [2x]^2 and whatnot are used. These type of questions also include incidences of "sparingly soluble".

I know my explanation is bad, but somewhat correct?

That's not exactly the question I was asking. [2OH-] has no physical meaning. It's senseless. What is it? The concentration of twice OH- ions? Twice the concentration of OH- ions? Why is the factor of two there? A solubility product constant is simply the equilibrium expression that characterizes the process of dissolution of a solid substance. Any equilibrium expression is [products]/[reactants]. Reactant in this case is a pure substance in a condensed state, which has an activity of one. (the denominator doesn't just disappear - if you weren't taught activity or just brushed over it, now might be a good time to wrap your head around the concept to understand this expression). Therefore, the equilibrium expression is simply [products]. Your products are one Ca2+ and two OH-. Therefore, Ksp = [Ca2+][OH-][OH-].

Here is a similar problem and explanation from the AAMC Chemistry Q-pack, and also a slightly more detailed explanation for the same problem. Try to remember you are relating Ksp and molar solubility.
that helped a lot, thank you

This sounds unusually complicated.

That's key. The solubility product constant (Ksp) is an equilibrium constant, so its expression is simply Ksp = [products] since the reactants are undissolved solid and liquid water which aren't included in the equation. For the reaction listed, Ksp = [Ca2+][OH-]^2. If you are calculating molar solubility from Ksp, set S = [Ca2+], which from stoichiometry means 2S = [OH-], resulting in Ksp = 4S^3.

Got it, thank you

That's not exactly the question I was asking. [2OH-] has no physical meaning. It's senseless. What is it? The concentration of twice OH- ions? Twice the concentration of OH- ions? Why is the factor of two there? A solubility product constant is simply the equilibrium expression that characterizes the process of dissolution of a solid substance. Any equilibrium expression is [products]/[reactants]. Reactant in this case is a pure substance in a condensed state, which has an activity of one. (the denominator doesn't just disappear - if you weren't taught activity or just brushed over it, now might be a good time to wrap your head around the concept to understand this expression). Therefore, the equilibrium expression is simply [products]. Your products are one Ca2+ and two OH-. Therefore, Ksp = [Ca2+][OH-][OH-].

Thanks!