Solubility KSP

[Farcus]

Ok

for example
K2CO3 (s) -> 2 K + (aq) + CO3 2- (aq)

ok I know KSP = product^coefficient so it would be [k]^2 * [CO3], but why is it sometimes (2s)^2 * s? I know S is just substituting the concentration but why the 2 in front of the s? What is the point of this? I am totally confused and been reading my book but doesn't seem to be helping.

---

Members don't see this ad.

 

[PrinceAli2786]

yea I always wondered that too, and when exactly do we use the 2s part and when do we not, some please let us know! lol thanks

 

[unsung]

If s = [CO3^2-], then [K+] = 2s. Why? Because for every molecule of K2CO3 that dissolves, we get two K+ ions and one CO3^2- ion. This means for every mole of K2CO3 that dissolves, we get 2 moles of K+ and 1 mole of CO3^2-. This means the concentration of K+ ions should be twice the concentration of CO3^2- ions. This is why you plug in (2s)^2 * (s).

 

[Farcus]

So let me get this straight

(2s)^2 * s

the 2 in (2s) is the number of ions there are in one unit, the ^*2, is what you get from the coefficient.

 

[engineeredout]

So let me get this straight

(2s)^2 * s

the 2 in (2s) is the number of ions there are in one unit, the ^*2, is what you get from the coefficient.

Yeah thats it. NaCl -> [1*s][1*s] = s^2

 

[engineeredout]

As a side note, when writing the actual equilibrium expression you don't include the coefficients before the Ss.

Ex: for K2CO3 its Ksp = [k+]^2 [CO3-2]. Only plug in the 2s when calculating the actual solubility and setting the equation equal to the Ksp constant given.