Chem 2

what is the solubility of silver carbonate in water in 25 degrees celcius if Ksp=8.4X10-12? i don't know what to do

Write the equation.
Ag2CO3(s) ==> 2Ag^+ + CO3^=

Ksp = (Ag^+2)(CO3^=)
Let x = solubility of Ag2CO3. At equilibrium,(Ag^+) = 2x; (CO3^=) = x. Plug those into the Ksp expression and solve for x. The answer comes out in mols/L. If you want grams, then M x molar mass Ag2CO3 = grams Ag2CO3.
Post your work if you get stuck.

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asked by joe
  1. Ag2CO3= 2Ag^1+CO3^-2
    Ksp=[Ag^+1]^2[CO3^-2]
    . =(2x)^2(x)
    8.4*10^-12=4x^3
    8.4*10^-12/4= 4x^3/4
    Cube root of 2.1*10^-12 =x
    0.0001280
    1.28*10^-4M=x

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