Hi guys, I'm working on some questions for a chemistry report and I honestly have no clue how to do it
Hopefully someone here can help

2a. In your notebook, write the dissociation equation for calcium hydroxide.
2b. Write the equilibrium constant expression (Ksp) for this equation.
2c. If the hydroxide ion concentration of a calcium hydroxide solution was found to be 0.004mol/L, what would be the concentration of the calcium ion?

I currently got
2A. Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
2B. Ksp=6.5x10-6

But no clue with 2C... help would be greatly appreciated! :)

2a. That looks correct to me.
2b. That also looks correct to me.
2c. You need the definition of Ksp

K_{sp} = [Ca^{2+}][OH^-]^2

where the "[]" operators mean the "concentration of"

You have the following:

6.5 \times 10^{-6} \,\frac {mol^3}{L^3} = [Ca^{2+}]\,\times\, (0.004)^2\, \frac {mol^2}{L^2}

All you have to do is solve that equation for [Ca^{2+}], the concentration of Calcium ions.

Hi Perito, thanks for the fast reply.

Can you please explain why the 0.004mol/L isn't squared due to there being 2 moles of OH?

Thanks

Quite right. I was sleepy last night so I've fixed my post.

Since you figured that out, I'm curious as to why you need our help.

Perito, I think there is a simpler way of looking at this. It takes 2 moles of OH for every mole of Ca. Thus, if the hydroxide ion concentration is 0.004mol/L then the calcium concentration would be half of that. 0.002mol/L

Perito, I think there is a simpler way of looking at this. It takes 2 moles of OH for every mole of Ca. Thus, if the hydroxide ion concentration is 0.004mol/L then the calcium concentration would be half of that. 0.002mol/L

That may be true in this situation, but in general, it's not. It is only true if you put Ca(OH)2 in pure water. If there's another source of hydroxide, the Ca(2+) concentration will drop.

I see, like if the water was self ionizing.