No. 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.
What mass of Ag2CO3 would be found in 1.4 L of a saturated solution if the Ksp of Ag2CO3 is 8.2 x 10-12?
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?
2 answers
Can u solve for y