Question

You have a solution that is 0.25M Ca2+ and .18M Ag+. You begin to add CO32- (carbonate ion) to form CaCO3 , and Ag2CO3. Which will precipitate first? Second?

What is the concentration of the first ion at the point when the second begins to precipitate?

What percentage of the first ion has been removed?

There three questions here.

Answer 1

[Ag+]=0.18 and [Ca+2]=0.250

Ksp Ag2CO3 = 8.1 * 10^(-12)

Ksp CaCO3 = 6* 10^(-9)

Now lead nitrate (CO3)2 -    is added to the solution.. thus forming compounds Ag2CO3 and CaCO3..

Among the above the one which requires lesser cooncentration of (CO3)2- will precipitate first

therefore (CO3)2- required for Ag2CO3 to precipitate =

[Ag+][(CO)32-]=Ksp(Ag2CO3), therefore [(CO3)2-]= 8.1*10^-12/0.18 =     4.5*10^-(11) M

similarly [(CO3)2-] required for CaCO3 to precipitate is Ksp(CaCO3)/[Ca+2]=6*10^(-9)/0.25 =     2.4*10^(-8) M

clearly the concentration of carbonate(CO3)2- required to precipitate Ag2CO3 is lesser .. therefore Ag2CO3 will get precipitated first.. when the concentration of carbonate ( CO3)2- reaches 2.4*10^(-8) M ,both CaCO3 and Ag2CO3 precipitate..

Hence Ag2CO3 will precipitate first

when second ion starts precipitating (CO3)2- = 2.4 * 10 ^ (-8)

hence [Ag+] = Ksp(Ag2CO3) / (CO3)2- = 3.375 * 10 ^ (-4)

percentage = ( 0.25 -   3.375* 10 ^(-4)    )   / 0.25    * 100 =     99.865 %

ans ) 99.865 %