At equilibrium, the concentration of calcium ions in a calcium chromate (CaCrO4) solution is 1 x 10^–4 mol/L at 25°C. In the same solution, the concentration of chromate ions is also 1 x 10^–4 mol/L. What is the Ksp of calcium chromate? Show your work.

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- vera_ewing

- katieb

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- vera_ewing

@aaronq Please help!

- welshfella

sorry I've forgotten this stuff

- aaronq

Write the dissociation, then it's corresponding equilibrium equation (which is the Ksp equation). Plug in your values and solve

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## More answers

- vera_ewing

What's the dissociation and corresponding equilibrium equation?

- aaronq

the dissociation of CaCrO4 (in water)

- aaronq

and by that i mean a chemical equation

- vera_ewing

Okay, so what does the chemical equation look like? Can you set it up and then I solve?

- aaronq

nah, you have to try it

- vera_ewing

I don't know how :( Can you maybe explain it?

- aaronq

CaCrO4 is an ionic compound, right? what ions make up CaCrO4?

- vera_ewing

Not sure. It's covalent and ionic though, right?

- aaronq

it's ionic only. You should be able to identify the ions here, if you can't look the compound up on wikipedia and read a little about it

- welshfella

what i do remember is that Ca CrO4 disassociates in water to
[Ca++] and [Cr2O4 --]

- welshfella

and sp means solubility product
so i guess Ksp is the product of Ca++ ions and CrO4-- ions

- vera_ewing

Yep, I figured it out :) thanks

- welshfella

so that 10-4 8 10^-4 = 10^-8

- welshfella

oh Ok wd

- Photon336

CaCrO4 ----> Ca^2+ + CrO4^2-
Ksp = [Ca^2+][CrO4^2-]
Ksp = [x][x]
Ksp = concentration of ions dissolved, like any K it's an equilibrium value, and it is temperature dependent.
I'm assuming that the equilibrium is already reached and that the solution is saturated. B.c from my understanding that's what Ksp is.
(Qsp = Ksp)
Their concentrations are equal for both ions in the problem.
So I think it's
(1x10^-4)(1x10^-4) = Ksp. 10^-8

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