help on a question plz

[afatasstank]

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Ca(OH)2 has Ksp = 10-6. If 0.1 mol of Ca(OH)2 was added to 1 L of pure water, what would be the approximate pH of the resulting solution?

the answer is 12, can u plz show and explain how to do it?
thanks

 

[Akarat]

Not sure how to find the cubed part without a calculator, but anyway:

First off, 0.1 mol of Ca(OH)2 is placed into 1 L of water. The Ksp = 1.0 x 10^-6, which is SO low that it doesn't matter. Therefore, we can ignore the molar concentration of Ca(OH)2.



R: Ca(OH)2 <---> Ca2+ + 2OH-
I: ******........... 0 ........ 0
C: ******......... +x ...... +2x
E: ******......... +x ...... +2x

* = Don't know, don't care

Ksp = (x)(2x)^2

1.0 x 10^-6 = 4x^3

x = 0.0063

Going back to the OH-, there's 2 mol of OH- that dissociate into solution for every mol of Ca 2+ that dissociates into solution.

2x => 2(0.0063) = 0.0126 mol OH-

Since the solution has a volume of 1 L, the molar concentration is
[OH-] = 0.0126

pOH = -log[OH-]
pOH = -log(0.0126)
pOH = 1.899 which is approximately equal to 2

pH = 14 - pOH
pH = 14 - 2
pH = 12

 

[afatasstank]

thank you very much for solving it akarat,
if you dont mind though, could u elaborate on a couple of things?
1st, when u say the ksp is so low, its so low relative to what? in other words how do u know when its low and negligible?

2nd, that ^ being asked, what if it wasnt so low and not negligible,

and finally, what is this RICE trick?

im sorry for this bombardment of questions after u already answered my original one beautifully, but it would really be helpful. thanks again.

 

[Akarat]

thank you very much for solving it akarat,
if you dont mind though, could u elaborate on a couple of things?
1st, when u say the ksp is so low, its so low relative to what? in other words how do u know when its low and negligible?

2nd, that ^ being asked, what if it wasnt so low and not negligible,

and finally, what is this RICE trick?

im sorry for this bombardment of questions after u already answered my original one beautifully, but it would really be helpful. thanks again.

As a general rule, when you're given a Ksp, it's probably going to be the solubility product of a mostly insoluble salt, such as Ca(OH)2. Since the salt is mostly insoluble, the amount that dissociates into solution is so low that it is insignificant relative to the original concentration.

If you subtract 2x (2x = 0.0126) from 0.1, you get 0.0874, which is relatively pretty small.

If the Ksp was not small, then you would probably have to use the quadratic equation to solve for x. Essentially, you would treat it like a weak acid equilibrium problem.

The RICE trick is just a table to keep track of your work when you're doing equilibrium dissociation problems. Reaction, Initial concentrations, Change, and Equilibrium concentrations. There's variations, some people called it an "ICE" table, but it's the same.

Lastly, if you were to set up the RICE table in such a way as not ignore the initial concentration of Ca(OH)2, you would get roughly the same pH.

Just to prove it:

R: Ca(OH)2 <---> Ca2+ + 2OH-
I: .... 0.1 ........... 0 ........ 0
C: 0.1 - x ......... +x ...... +2x
E: 0.1 - x ......... +x ...... +2x

Ksp = (x)(2x)^2 / (0.1 - x)
1.0 x 10^-6 = 4x^3 / (0.1 - x)

I used a TI - 89, but x comes out to be 0.002896

2x = 2(0.002896) = 0.005792

pOH = -log(0.005792)
pOH = 2.237 which is approximately 2
=> pH = 12

 

[afatasstank]

thanks so much! i owe you one.

 

[Akarat]

If you subtract 2x (2x = 0.0126) from 0.1, you get 0.0874, which is relatively pretty small.

Small correction, it's actually 0.1 - 0.0063, and not 0.1 - 0.0126. This is because the equilibrium concentration for Ca(OH)2 is 0.1 - x, if you didn't ignore the change in concentration for Ca(OH)2 in the first place. Moreover, 1 mol of Ca(OH)2 yields 2 mol of OH-.

Regardless, everything else is the same.