Calculate the [Ag+] and [CO3 2-] in the solution at equilibrium when 6 mL of 0.1 M Na2CO3 are mixed with 3 mL of 0.5 M AgNO3?
The solubility product constant of Ag2CO3 is 8.1x10^-12 at 25'C. By constructing the ICE table, calculate the [Ag+] and [CO32-] in the solution at equilibrium when 6 mL of 0.1 M Na2CO3 are mixed with 3 mL of 0.5 M AgNO3.
- hcbiochemLv 72 months agoFavourite answer
Moles Ag+ initially present = 0.003 L X 0.5 mol/L = 0.015 mol Ag+
Moles CO32- initially present = 0.006 X 0.1 mol/L = 6X10^-4 mol
Since CO32- is the limiting reactant, initially assume that all of it precipitates.
Moles Ag+ precipitated = 1.2X10^-3 mol
Moles Ag+ remaining in solution = 0.0138 mol
[Ag+] remaining = 0.0138 mol / 0.009 L = 1.53 M
Ksp = [Ag+]^2 [CO32-] = 8.1X10^-12
8.1X10^-12 = 1.53^2 [CO32-]
[CO32-] = 3.4X10^-12 M
I know...I didn't use an ICE table, but you can format this into one. As I did, assume that all of the CO32- precipitated as Ag2CO3, and then let [CO32-] = x, [Ag+] = 1.53+2x. Assume that 2x<<1.53, and calculate [CO32-]