Can someone please help me understand this equilibrium constant problem

If the solubility of Li2CO3= 0.15 moles/liter, what is its Ksp at this temperature?
so I know the equilibrium is Li2CO3 <--> 2Li + CO3 and based off of the equation for K it would be
K=[Li]^2[CO3]

so I thought I would solve it this way
K=[.15]^2[.15]

BUT its actually
K=[2x.15]^2[.15]
K=[.30]^2[.15]

why do you multiply by 2 AND square it? do you always do this? please help! test tomorrow!!! thanks :)

Because you don't know the molarity of Li.

All you know is that when Li2CO3 is broken down there are two moles of Li created.

Since Xmoles of Li2CO3 creates X moles of Li: K = [2x][x], and since you have a coefficient of two for Li you need to square Li, so K = [2x]^2[x].

If the concentration of Li was known you would not need to multiply the molarity Li by the coefficient, but since you only know the molarity of Li2CO3 you have to multiply by the coefficient.

When you're drawing your ICE (Initial, Change, Equilibrium) table, remember that for each carbon trioxide ion that dissociates, two lithium ions dissociate.

.....Li2CO3......[Li].......[CO3]
.I ````Initial```````````0````````````0`
C`````-x````````````+2x`````````+x
E``Initial-x`````````2x```````````x

(I put the Li2CO3 in only for the sake of a balance ICE table, it has no effect on the constant.)

Ksp = [Li]^2 * [CO3]
But, [Li] = 2x and [CO3] = x, so
Ksp = (2x)^2 * (x)

You multiply it by two because you're making TWO moles of Li+ but only ONE mole of CO3-. That's why the formula is Li2CO3 ---> TWO Li+ plus ONE CO3-.

Therefore there are .15 mol Li2CO3, .30 mol Li+, and .15 mol CO3-

Yes you always do this. :)