Question

When excess Ag2CO3(s) is shaken with 1.00 L of 0.200 M K2CO3 it is determined that 6.00 x 10ˉ⁶ moles of Ag2CO3 dissolves. Calculate the solubility product of Ag2CO3. Please provide step by step. Thank you.

Answer 1

solubility of Ag2CO3 is given by the following equation

Ag2CO3(s) <-----> 2Ag+(aq) + CO3^2-(aq)

Ksp = [ Ag+ ]^2 [ CO3^2-]

Now ,

the concentration of K2CO3 = 0.2M

Concentration of CO3^2- = 0.2M

Volume = 1L

moles of CO3^2- = 0.2mol

Dissolved moles of Ag2CO3 = 6.00×10^-6

Dissolved moles of Ag+ = 1.20×10^-5mol

Total mole of CO3^2- = 0.2 + 0.000006mole = 0.200006mol~0.2mol

[ Ag+ ] =1.2×10^-5M

[ CO3^2-] = 0.2M

Therefore,

Ksp = (1.2×10^-5)^2(0.2)

= 2.9×10^-11