Question #207a8

Like always start with what you know.

You know that silver carbonate#"Ag"_2"CO"_3#, is an insoluble ionic compound that has a solubility product constant equal to #8.1 * 10^(-12)#.

You know that the very small amounts that do dissociate in aqueous solution will form silver cations, #"Ag"^(+)#, and carbonate anions, #"CO"_3^(2-)#.

This dissociation is an equilibrium reaction, which implies that the indissolved solid will be at equilibrium with its dissolved ions in accordance to the value of the solubility product constant.

You also know that equilibrium reactions are governed by Le Chatelier's Principle, which states that any equilibrium will act in such a manner as to counteract any disturbance that affects it.

In your case, this disturbance is actually the presence in solution of the carbonate anions.

To counter this disturbance, the equilibrium will shift to the left, meaning that less ions will dissociate from the solid silver carbonate - this is known as the common ion effect.

Use an ICE table to help set this up - the initial concentration of the carbonate ions will be that given to you in the problem

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

#color(purple)("I")" " " " -" " " " " " " " " " " " " 0" " " " " " " " " "0.025#
#color(purple)("C")" " " " -" " " " " " " " " "(+color(red)(2)s)" " " " " " "(+s)#
#color(purple)("C")" " " " -" " " " " " " " " " " "(color(red)(2)s)" " " " " " "(0.025+s)#

By definition, the solubility product constant will be

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

#K_(sp) = (color(red)(2)s)^color(red)(2) * (0.025 + s) = 4s^3 + 0.1s^2#

The cubic equation

#4s^3 + 0.1s^2 - 8.1 * 10^(-12) = 0#

will produce three values for #s#, out of which only one will be positive - you need #s# to be positive because it represents concentration, which has no chemical and physical meaning as a negative value

#s = 9.0 * 10^(-6)"M"#

This means that the concentration of silver cations in this solution will be

#["Ag"^(+)] = color(red)(2) * s = 2 * 9.0 * 10^(-6) = color(green)(1.8 * 10^(-5)"M")#