Posted by molar solubility on Saturday, July 30, 2011 at 7:44pm.
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?
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.
what is Ksp expression of mercury(I) chloride Hg2Cl2 in terms of molar solubility, s?
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
- chemistry - DrBob222, Saturday, July 30, 2011 at 10:50pm
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-) = 8.1E-12
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) = 8.1E-12.
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)=8.1E-12 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) ==> Hg22+ + 2Cl^-
Ksp = (Hg22+)(Cl^-)^2
So Hg22+ = x and Cl^- is 2x. Substituting gives 4s^3
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