[FthkuMan]

Hi guys, I have the following question:

Ca(OH)2 was added to a 1 liter solution containing 0.1M NH3 (Kb=1.8*10^[-5])
How many grams of Ca(OH)2 will dissolve in the water?
Ksp = 5*10^[-7]

A. 0.10
B. 0.17
C. 0.35
D. 1.2

What I did was calculate the number of OH- already existing in the solution, substracted it from the solubility and then muliplied by the molar mass of Ca(OH)2, I came up with roughly 0.33, but it just doesn't seem like that would be the right way.

Basically what I did was this:

NH3 + H2o -> OH- + NH4+
0.1-x               x         x
Kb= 1.8*10^[-5]
x=1.34*10^[-3]

Ksp = 5*10^[-7], divide by 4 and find the cubic root
S= 5*10^[-3]
2s (for OH) = 0.01

0.01 - 1.34*10^[-3] = 8.66*10^[-3]

Then divide that by 2 and multiply by 74.
I'm sure this isn't the right way, any help would be appreciated!

[Borek]

TBH, no idea what you did.

What was the initial pH? How does the presence of OH^- affect amount of Ca(OH)₂ that can dissolve?

This can get tricky, but you can start just by assuming initial pH doesn't change (that is, concentration of OH^- is constant).

[FthkuMan]

TBH, no idea what you did.

What was the initial pH? How does the presence of OH^- affect amount of Ca(OH)₂ that can dissolve?

This can get tricky, but you can start just by assuming initial pH doesn't change (that is, concentration of OH^- is constant).

OK, so I calculated the [OH] and it's 1.34*10^[-3]M. How do I use that to find the number of grams of the salt that were dissolved?
Simply multiply the [OH] by 74? (the molar mass of the salt)

[Borek]

Simply multiply the [OH] by 74? (the molar mass of the salt)

No, that's completely random (more or less about as random as your initial attempt).

I already gave you a hint: assume OH^- doesn't change during dissolution. What is the highest concentration of Ca²⁺ that can survive in the solution before the precipitation?

[FthkuMan]

Simply multiply the [OH] by 74? (the molar mass of the salt)

No, that's completely random (more or less about as random as your initial attempt).

I already gave you a hint: assume OH^- doesn't change during dissolution. What is the highest concentration of Ca²⁺ that can survive in the solution before the precipitation?

EDIT: well,
5*10^-7=[Ca²⁺][OH^-]²
5*10^-7=[Ca²⁺]*(1.34*10^-3)²
[Ca²⁺] = 0.278

Doesn't make much sense to me though.

[Borek]

You should check if the assumption that [OH^-] didn't change holds. It probably doesn't, so the problem gets tricky (as adding OH^- from the dissolution shifts the ammonia dissociation equilibrium). It may even happen that it is ammonia dissociation that can be safely ignored.

Sorry to say that, but if this approach doesn't make sense to you, you should reread your book, especially the part about the common ion effect and how Ksp works.

[FthkuMan]

You should check if the assumption that [OH^-] didn't change holds. It probably doesn't, so the problem gets tricky (as adding OH^- from the dissolution shifts the ammonia dissociation equilibrium). It may even happen that it is ammonia dissociation that can be safely ignored.

Sorry to say that, but if this approach doesn't make sense to you, you should reread your book, especially the part about the common ion effect and how Ksp works.

It wasn't the approach that didn't make sense, rather the result. I'm just not sure how to go about solving this. Again, I would appreciate any help.

[Borek]

I already told you to check if the pH has changed. Did it? By how much?

[FthkuMan]

I already told you to check if the pH has changed. Did it? By how much?

That's the part I'm not sure how to go about.
OK, so calculating the initial pH of the solution it comes up to 11.12.
How do I then factor in the added [OH]? do I add the solubility, which I can get from the Ksp, to the existing number of moles of [OH]?
But, seeing as this is a basic solution, the solubility will decrease.

[Borek]

As I said, this can easily get tricky.

But let's forget for a moment about changes in solubility. How much OH^- is produced in the dissolution? Assuming it just adds to the already existing concentration of OH^-, what is the new pH?

[FthkuMan]

As I said, this can easily get tricky.

But let's forget for a moment about changes in solubility. How much OH^- is produced in the dissolution? Assuming it just adds to the already existing concentration of OH^-, what is the new pH?

It would add 0.01M of OH^-, the new concentration would be 0.01134 and so the new pH would be 12.05.

EDIT: There is a chance the initial concentration of NH3 might have been 1.0M and not 0.1M.. but only maybe. 

[docnet]

 

[Borek]

[Ca²⁺] = 0.278

It would add 0.01M of OH^-

Both those numbers can't be right at the same time.

[FthkuMan]

[Ca²⁺] = 0.278

It would add 0.01M of OH^-

Both those numbers can't be right at the same time.

Yes, that was before I calculated the [OH^- the way you told me. It would now be [Ca²⁺] = 3.89*10^[-3]
Going by this solubility, the grams dissolved would be 0.28, which is neither of the answers.