Precipitation Question

[marymatthews]

---

Members do not see this ad.

Is this the correct solution to the following question please?

0.0001 M solution of Mg(NO3)2, and NaOH is added to a final conc of 0.001 M. Did a ppt form? (Ksp Mg(OH)2 = 1.6x10^-12)

Ksp = [Mg+2][OH-]^2
= (.0001 M)(0.001 M)^2 = 1x10^-10
Yes, ppt forms

My question is, do you multiply the Molarity of OH- by 2 though?
so it would be
= (.0001 M)(2 x 0.001 M)^2 = 1x10^-10

 

[lorenzomicron]

no. the coefficient determines the power of the concentrations in the rate equation.

 

[marymatthews]

So then why do you multiply by 2 other times?

 

[lorenzomicron]

what other times?

if the equation says 1 A + 2 B = 1 C
then it indicated 1 molecule of A and 2 molecules of B must collide into the intermediate in order for the product to form. the kinetic theory says this results in relying on the square of the concentrations, such that
K= [A]^1*^2...

or did i miss something?

 

[marymatthews]

For example:
BiI3 <=> Bi+3 + 3I-1
Ksp = [Bi+3][I-1]^3
Ksp = (x)(3x)^3
Ksp = 27x^4

 

[ilovemcat]

Is this the correct solution to the following question please?

0.0001 M solution of Mg(NO3)2, and NaOH is added to a final conc of 0.001 M. Did a ppt form? (Ksp Mg(OH)2 = 1.6x10^-12)

Ksp = [Mg+2][OH-]^2
= (.0001 M)(0.001 M)^2 = 1x10^-10
Yes, ppt forms

My question is, do you multiply the Molarity of OH- by 2 though?
so it would be
= (.0001 M)(2 x 0.001 M)^2 = 1x10^-10

Yeah, you're right - although I didn't attempt the calculation.


Here the gave you the concentration of both ions:

[Mg] = 1x10^-4
[OH] = 1x10-3

Because 0.001 represents 1 mole of OH (in NaOH) but there are 2 moles of OH in Mg(OH)2, you would multiply that concentration by 2. Then you'd plug that formula in and solve for the Qsp ([Mg][2 x OH]^2]).

If Qsp is LESS than Ksp, then no precipitate formed.
However, if Qsp is greater than Ksp then precipitate did form.

 

[marymatthews]

So it should be:
= (.0001 M)(2 x 0.001 M)^2 = 1x10^-10

 

[ilovemcat]

So it should be:
= (.0001 M)(2 x 0.001 M)^2 = 1x10^-10

Yes, but the calculation should come out to 4x10^-10 M^3.

 

[golgiapparatus88]

If you are running out of time on the test and see this question, can't you just realize that a really small Ksp value like 1.6x10^-12 means that precipitation will occur?

 

[ilovemcat]

If you are running out of time on the test and see this question, can't you just realize that a really small Ksp value like 1.6x10^-12 means that precipitation will occur?

I think you're confusing the idea of Ksp. Ksp = Solubility Product which is just another form of the equilibrium constant for solutes dissolving in water. If you attain a Qsp greater than Ksp, then precipitate will form but if it's less there won't be any precipitate.

Another way of explaining is this is say you're adding some solid to water (your solvent). You keep adding a little bit of the solid (ie, Mg(OH)2) and each time it dissolves. However, you eventually reach a point where the solution becomes saturated with solute and no more can dissolve. That's what Ksp is for. If you exceed the Ksp value, precipitate will form.

 

[golgiapparatus88]

That's true for this case but if there is a question that says "give the order of which will form a precipitate first to last" then it would be the smallest Ksp up to the highest Ksp. There's a really good example in a TBR gen chem passage.

 

[Rosalindbungs]

Ive always had trouble with equations that involve calculations and setting the equations up... for this one... how did you get the final equation Mg(OH)2=Mg2+ + OH-? Was it just how you interpreted the qeustion?

My line of logic was Mg(NO3)2 is soluble so it dissociates into Mg2+ and No3- ions... then you add NaOH, which is in Na+ and OH- form... and then I got stuck...

What was your guys's reasoning behind how you got the final concentration calculations?

 

[MediCynical]

So it should be:
= (.0001 M)(2 x 0.001 M)^2 = 1x10^-10

Yes, but the calculation should come out to 4x10^-10 M^3.

No, this is incorrect. The above example of

For example:
BiI3 <=> Bi+3 + 3I-1
Ksp = [Bi+3][I-1]^3
Ksp = (x)(3x)^3
Ksp = 27x^4

is correct, because although the ratio of the I- ion to the Bi ion is 3, it can only be said to equal 3x because Bi and 3I are in solution in the same concentrations - they came from the same species which was put in a solution at a given concentration of x.

In the example from the original post, Mg+2 and OH- are in solution in different concentrations - they did not come from the same species being dissolved in solution.

You had it correct before:
Ksp = [Mg+2][OH-]^2
= (.0001 M)(0.001 M)^2 = 1x10^-10

 

[MediCynical]

Ive always had trouble with equations that involve calculations and setting the equations up... for this one... how did you get the final equation Mg(OH)2=Mg2+ + OH-? Was it just how you interpreted the qeustion?

My line of logic was Mg(NO3)2 is soluble so it dissociates into Mg2+ and No3- ions... then you add NaOH, which is in Na+ and OH- form... and then I got stuck...

What was your guys's reasoning behind how you got the final concentration calculations?

you're going in the right direction, just think of the charges of the ions. When both Mg(NO3)2 and NaOH are dissolved in solution, you end up with two positive ions and two negative ions - Na+ & Mg+2, and NO3- & OH-. Since Mg has a charge of 2+, it will need two - charges to be neutral. It tells you the Ksp of Mg(OH)2, so that's what we're looking for. Mg(OH)2 comes from one of the Mg+ ions of Mg(NO3)2 and two of the OH- ions from NaOH.

I don't know if that's what you were looking for...