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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@aaronq Please help!
sorry I've forgotten this stuff
Write the dissociation, then it's corresponding equilibrium equation (which is the Ksp equation). Plug in your values and solve
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What's the dissociation and corresponding equilibrium equation?
the dissociation of CaCrO4 (in water)
and by that i mean a chemical equation
Okay, so what does the chemical equation look like? Can you set it up and then I solve?
nah, you have to try it
I don't know how :( Can you maybe explain it?
CaCrO4 is an ionic compound, right? what ions make up CaCrO4?
Not sure. It's covalent and ionic though, right?
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
what i do remember is that Ca CrO4 disassociates in water to
[Ca++] and [Cr2O4 --]
and sp means solubility product
so i guess Ksp is the product of Ca++ ions and CrO4-- ions
Yep, I figured it out :) thanks
so that 10-4 8 10^-4 = 10^-8
oh Ok wd
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