[sundrops]

the solubility of Ca(OH)₂ is 0.0510g/100mL H₂O at 25 C.

calculate [Ca2+] and [OH-] in standard solution of Ca(OH)2.

Calculate the value of Ksp.
I know Ksp = [Ca2+][OH-]²
but I cannot find Ksp before I determine the concentrations of the products. And I'm having a little bit of trouble doing that.

Here's what I have done so far:

(0.0510g/100mL H2O) * (1000mL H2O / 74.096g/mol a(OH)2) = 0.00688296mol/L Ca(OH)2  and water

my question is - what do I do with this water? do I simply disregard it and continue with the problem?

if I just ignore the water, I can form my ice table and easily find that the concentrations are:
[Ca2+] = 0.0068829 M
[OH-] = 1.8950055E-4 M

Ksp = 1.30431E-6 M

does that look ok to you guys?

[Borek]

Calculate the value of Ksp.
I know Ksp = [Ca2+][OH-]²
but I cannot find Ksp before I determine the concentrations of the products. And I'm having a little bit of trouble doing that.

Here's what I have done so far:

(0.0510g/100mL H2O) * (1000mL H2O / 74.096g/mol a(OH)2) = 0.00688296mol/L Ca(OH)2  and water

So far so good.

my question is - what do I do with this water? do I simply disregard it and continue with the problem?

I have no idea why water makes any difficulty here. You have a solution of Ca(OH)₂ and you assume what was dissolved is completely dissociated. Water is just a solvent. You have correctly calculated concentration of Ca(OH)₂.

if I just ignore the water, I can form my ice table and easily find that the concentrations are:
[Ca2+] = 0.0068829 M
[OH-] = 1.8950055E-4 M

What do you use ICE table for? Assume all calcium hydroxide is dissociated.

[AWK]

[OH-] = 1.8950055E-4 M

?
[OH-] = 2[Ca2+]
You can neglect {OH-} from water because [OH-] from water is ~100000 times less than [OH-] from Ca(OH)2

[sundrops]

wouldn't [OH-] = 2[Ca2+]² ?

alright guys - thank you very much! ... just one more question...

now if I wanted to neutralize 10mL of this solution with 0.0150M HCl how would I determine the volume HCl required?

Here's what I have done thus far:

now a neutralized solution would have a Ksp of 7 right?
as it stands now, the Ksp of HCl is 4.7375E-5

how do I go about doing a neutralzation problem? I'm not really sure how to start going about neutralizing 10 mL   sorry guys - but I'd really appreciate your input.

[Borek]

wouldn't [OH-] = 2[Ca2+]² ?

No. Where did yiu get the square from?

now if I wanted to neutralize 10mL of this solution with 0.0150M HCl how would I determine the volume HCl required?

Simple stoichiometric problem. Write down the reaction equation and you are almost done.

now a neutralized solution would have a Ksp of 7 right?
as it stands now, the Ksp of HCl is 4.7375E-5

No. Not everything in chemistry is called Ksp. Solution doesn't have Ksp, HCl doesn't have Ksp.

[sundrops]

oh ok - makes sense, thanks borek  

ok so here's how far I've gotten with this stoicheometry reaction stuff:

0.0068296mol/L Ca(OH)2 * 0.01L *0.0150mol/L HCl * 36.458g/mol HCl * 74.096g/mol Ca(OH)2 = 0.002767 g²/L

how do i get it in just litres? Is it alright the way I did it?

[Borek]

Forget about grams, all you have to do is to calculate number of HCl moles necessary for neutralization, then calculate volume of the solution that contains this number of moles. n=CV stuff.

Where did you get g²/L from? Seems you like everything squared

[sundrops]

yeah i guess I do   haha

ok, now how do I know when my solution becomes neutralized?
is it just a ration thing? in that case there is a 1:1 ration between Ca(OH)2 and HCl

so then I move onto my C1V1=C2V2
(0.0068296)(0.01) / 0.0150 = 0.0045531 L

does that look ok?

sorry if it seems like I need to be led by the hand.... I really trying to understand this solubility stuff, just takes me a little longer.  

[sundrops]

* ration = ratio
TYPO - my bad  

[Borek]

ok, now how do I know when my solution becomes neutralized?
is it just a ration thing? in that case there is a 1:1 ration between Ca(OH)2 and HCl

No. Solution is neutralized when all OH^- ions reacted with H⁺.

[sundrops]

okay.

so how about if I do this:

10mL Ca(OH)2 * 1L/1000mL * 0.0136592 mol OH-/L NaOH = 1.36592E-4 mol OH-

then since H+ ions react with OH- ions in a 1:2 ratio we need 6.8296E-5 mol of H+ ions to neutralize the reaction. so,

V * 0.0150 mol H+ / L = 6.8296E-5
V = 0.00455 L of HCl is required to neutralize the reaction.

Does this look better?

*crosses fingers*

[Borek]

0.0136592 mol

Where did you get this value from?

then since H+ ions react with OH- ions in a 1:2 ratio

No. They react 1:1. Ca(OH)₂ reacts 2:1 with HCl.

Does this look better?

Not for me. My result is twice as much.

*crosses fingers*

Didn't help.

[sundrops]

Quote: "Where did you get this value from?"

2mol OH for every mole Ca(OH)2

so I should try again with .oo688 instead of .013?

but the rest looks ok?