Question #32388

1 Answer
Sep 10, 2015

Answer:

For part (b) #s = 1.2 * 10^(-4)"M"#

Explanation:

I'll show you how to find the molar solubility of silver carboinate, #"Ag"_2"CO"_3#.

Silver carbonate is considered insoluble in aqueous solution, which means that it does not dissociate completely into silver cations, #"Ag"^(2+)#, and carbonate anions, #"CO"_3^(2-)#, when dissolved in water.

Actually, the fact that it doesn't dissociate completely is an understatement. When silver carbonate is placed in aqueous solution, an equilibrium reaction is established

#"Ag"_2"CO"_text(3(s]) rightleftharpoons 2"Ag"_text((aq])^(+) + "CO"_text(3(aq])^(2-)#

The solubility product constant, #K_(sp)#, essentially tells you the extent of dissociation.

The smaller the value of #K_(sp)#, the fewer ions will dissociate in solution. Of course, this implies a lower solubility, since most of the solid will remain undissolved.

To find the molar solubility of silver carbonate, use an ICE table

#"Ag"_2"CO"_text(3(s]) " "rightleftharpoons" " color(red)(2)"Ag"_text((aq])^(+)" " + " " "CO"_text(3(aq])^(2-)#

#color(purple)("I")" " " " - " " " " " " " " " " " " 0 " " " " " " " " " " 0#
#color(purple)("C")" " " " - " " " " " " " " " " (+color(red)(2)s) " " " " " " "(+s)#
#color(purple)("E")" " " " - " " " " " " " " " " " " color(red)(2)s " " " " " " " " " " s#

By definition, #K_(sp)# will be

#K_(sp) = ["Ag"^(+)]^color(red)(2) * ["CO"_3^(2-)]#

#K_(sp) = (2s)^2 * s = 4s^2 * s = 4s^3#

This means that #s#, which is the molar solubility of silver chloride, is equal to

#s = root(3)(K_(sp)/4) = root(3)( (6.2 * 10^(-12))/4) = color(green)(1.2 * 10^(-4)"M")#