chemistry
posted by molar solubility .
What is the molar solubility of silver carbonate Ag2CO3 at 25 C if it is dissolved in a 0.15M solution of silver nitrate, AgNO3?
Solubility constant:
silver carbonate = 8.1 *10^(12)
Answer: 3.6 *10^(10)
So... not sure exactly what to do here.
AgCO3 + AgNO3 ><
AgCO3 + Ag(+) + NO3() (?)
not sure what molarity has to do with this.
Any help would be appreciated.
Also
what is Ksp expression of mercury(I) chloride Hg2Cl2 in terms of molar solubility, s?
Answer: 4s^3
Why is this not s^2? I thought that mercury(I) was a diatomic and Cl2 had a 2 charge (and is also diatomic). That would mean
1 mercury(I)ion:1 chlorine ion or
s*s=s^2

The 0.15M has a lot to do with it. You have Ag^+ from two sources.
Ag2CO3 ==> 2Ag^+ + CO3^2 AND
AgNO3 ==> Ag^+ + NO3^
Ksp = (Ag^+)^2(CO3^2) = 8.1E12
If you want to go through it rigorously, then Ag^+ from Ag2CO3 is 2x and CO3^2 is x and Ag^+ from AgNO3 (it is soluble and 100% ionized) is 0.15. Substituting we get
(2x+0.15)^2(x) = 8.1E12.
The easy way to solve this is to recognize that the solubility of Ag2CO3 is small so 2x+0.15 is essentially equal to 0.15. The equation then becomes
(0.15)^2(x)=8.1E12 and solve for x. That gives you the CO3^2 which ALSO is Ag2CO3.
For your second question, note the formula is Hg2Cl2 and NOT HgCl2.(mercury(I) chloride is a dimer.) so
Hg2Cl2(s) ==> Hg2^{2+ + 2Cl^ Ksp = (Hg22+)(Cl^)^2 So Hg22+ = x and Cl^ is 2x. Substituting gives 4s^3}