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April 16, 2014

April 16, 2014

Posted by **molar solubility** on Saturday, July 30, 2011 at 7:44pm.

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

- chemistry -
**DrBob222**, Saturday, July 30, 2011 at 10:50pmThe 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) ==> Hg2^{2+ + 2Cl^- Ksp = (Hg22+)(Cl^-)^2 So Hg22+ = x and Cl^- is 2x. Substituting gives 4s^3 }

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