Gen chem Question (difficulty 2/10)

[liveoak]

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okay best friends. here is the question

the solubility of mg(OH)2 is 1.2 X 10-11 mol^3/l^3. What is the OH concentration in a sat solution?

So, the equation is mg(OH)2 -> mg2+ + 2OH

thus, you find "x" via (x)(2x)^2 = 1.2X 10^-11

then, once you have x...how do you apply it to find the concentration of OH?

is it 2x or 2x^2?

 

[MT Headed]

You defined X as the equilibrium concentration of Mg2+ in solution, in mol/L.
And you correctly used 2X as the equilibrium concentration of OH- in your equation.

(2x)^2 is an interesting math expression that can be used in an equilibrium constant calculation, but it has nothing to do with the concentration of OH. It doesn't even have the correct units.

 

[liveoak]

You defined X as the equilibrium concentration of Mg2+ in solution, in mol/L.
And you correctly used 2X as the equilibrium concentration of OH- in your equation.

(2x)^2 is an interesting math expression that can be used in an equilibrium constant calculation, but it has nothing to do with the concentration of OH. It doesn't even have the correct units.

Boom! thank you. let's get coffee later!

 

[SN1]

okay best friends. here is the question

the solubility of mg(OH)2 is 1.2 X 10-11 mol^3/l^3. What is the OH concentration in a sat solution?

So, the equation is mg(OH)2 -> mg2+ + 2OH

thus, you find "x" via (x)(2x)^2 = 1.2X 10^-11

then, once you have x...how do you apply it to find the concentration of OH?

is it 2x or 2x^2?

Equation:
mg(OH)2 -> mg2+ + 2OH

ksp = [mg2+][OH-]^2

mg(OH)2 releases twice that amount of OH.
molar solubility of mg(OH)2 is 1.2 x 10^-11(also molar concentration of mg2+) so molar concentration of OH is twice as that which is 2 x 1.2 x 10^-11

2.4 x 10^-11 M --------> ANSWER

You were given X in the problem

 

[MT Headed]

The original question is worded poorly. It claims to have a value for solubility, but its units are mol^3/L^3.

If it really was a solubility, its units would be g/L.
If it really was a molar solubility, its units would be mol/L.
If it really was a solubility-product constant, it would be unitless (K concentrations are measured in dimensionless "activities", which are thermodynamic quantities that numerically have the same value as concentrations).
The question as originally stated makes no sense.

I took a wild guess that the 1.2x10^-11 value was a solubility-product constant Ksp, the product of [Mg2+][OH-]^2, despite its incorrect units. This is further confirmed by the literature solubility-product value of Mg(OH)2 which is Ksp=1.5x10^-11. The solubility and molar solubility of Mg(OH)2 are nowhere near this order of magnitude.

I stand by my post #2, and believe post #4 is incorrect.

 

[SN1]

The original question is worded poorly. It claims to have a value for solubility, but its units are mol^3/L^3.

If it really was a solubility, its units would be g/L.
If it really was a molar solubility, its units would be mol/L.
If it really was a solubility-product constant, it would be unitless (K concentrations are measured in dimensionless "activities", which are thermodynamic quantities that numerically have the same value as concentrations).
The question as originally stated makes no sense.

I took a wild guess that the 1.2x10^-11 value was a solubility-product constant Ksp, the product of [Mg2+][OH-]^2, despite its incorrect units. This is further confirmed by the literature solubility-product value of Mg(OH)2 which is Ksp=1.5x10^-11. The solubility and molar solubility of Mg(OH)2 are nowhere near this order of magnitude.

I stand by my post #2, and believe post #4 is incorrect.

If 1.2 x 10^-11 is the Ksp then yes you are correct. But as you said, Ksp should NOT have a unit.

If indeed 1.2 x 10^-11 is the ksp then
ksp = [Mg+2] [OH-]^2
1.2 x 10^-11 = [x] [2x]^2
1.2 x 10^-11 = 4x^3
x = 1.44 x 10^-4--------> molar solubility

then concentration of OH- would just be twice as that which is 2.88 x 10^-4

However, I believe my original answer is correct IF 1.2 x 10^-11 was molar solubility of Mg(OH)2

 

[liveoak]

the question isn't worded correctly because i wasn't interested in the answer. just how to find the concentration of OH in the saturated solution.