Question #9e5cc
1 Answer
Explanation:
You know that when silver carbonate, a salt that is considered insoluble in water, is dissolved in water, the following dissociation equilibrium is established in aqueous solution.
#"Ag"_ 2"CO"_ (3(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "CO"_ (3(aq))^(2-)#
You also know that the molar solubility of silver carbonate,
This tells you that a saturated solution of silver carbonate will contain
In other words, you have
#["Ag"_ 2"CO"_ 3]_ "that dissociates" = s#
In your case, you know that
#["Ag"_ 2"CO"_ 3]_ "that dissociates" = s = 4.7 * 10^(-5)# #"M"#
Now, notice that every mole of silver carbonate that dissociates in aqueous solution produces
This means that the saturated solution of silver carbonate will contain
#["Ag"^(+)] = color(red)(2) xx s#
#["CO"_3^(2-)] = s#
You can thus say that at
#["Ag"^(+)] = color(red)(2) xx 4.7 * 10^(-5)color(white)(.)"M" = 9.4 * 10^(-5)color(white)(.)"M"#
#["CO"_3^(2-)] = 4.7 * 10^(-5)color(white)(.)"M"#
By definition, the solubility product constant for this dissociation equilibrium is equal to
#K_(sp) = ["Ag"^(+)]^color(red)(2) * ["CO"_3^(2-)]#
In your case, you will have--I'll leave the value without added units
#K_(sp) = (9.4 * 10^(-5))^color(red)(2) * (4.7 * 10^(-5))#
#color(darkgreen)(ul(color(black)(K_(sp) = 4.2 * 10^(-13))))#
The answer is rounded to two sig figs, the number of sig figs you have for the molar solubility of the salt.