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March 29, 2015

March 29, 2015

Posted by **Reindier** on Friday, April 18, 2008 at 8:11pm.

My Work:

SInce mass = number of mols x Molar mass, I first look for the molar mass of Ag2CO3, which is 275.75 g/mol. Multiplying .0000000000082 by 275.5 gives a value of 2.0 x 10^-9 g, which becomes the mass.

Is this correct?

- chemistry -
**DrBob222**, Friday, April 18, 2008 at 9:12pmNo. You haven't used Ksp at all except in a superficial manner. First you must determine the solubility of Ag2CO3. Use Ksp for that.

Ag2CO3 ==> 2Ag^+ + CO3^=

Ksp = (Ag^+)^2(CO3^=) = 8.2 x 10^-12

Solubility Ag2CO3 in mols/L = y

Then (Ag^+) = 2y

(CO3^=) = y

Now substitute those variables into Ksp expression.

(2y)^2(y) = 8.2 x 10^-12

solve for y which is (Ag2CO3)in mols/L.

Convert to grams by grams = mols x molar mass. That is the mass in 1 L.

Then that mass x 1.4 L/1.0 L = the mass of Ag2CO3 in 1.4 L. Post your work if you need additional assistance.

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