Question

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.

Answer 1

@aaronq Please help!

Answer 2

sorry I've forgotten this stuff

Answer 3

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

Answer 4

What's the dissociation and corresponding equilibrium equation?

Answer 5

the dissociation of CaCrO4 (in water)

Answer 6

and by that i mean a chemical equation

Answer 7

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

Answer 8

nah, you have to try it

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

what i do remember is that Ca CrO4 disassociates in water to

[Ca++] and [Cr2O4 --]

Answer 14

and sp means solubility product

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

Answer 15

Yep, I figured it out :) thanks

Answer 16

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

Answer 17

oh Ok wd

Answer 18

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