Question #9e5cc

1 Answer
Aug 3, 2017

Answer:

#K_(sp) = 4.2 * 10^(-13)#

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, #s#, in water at #25^@"C"# is equal to #"4.7 * 10^(-5)# #"M"#.

This tells you that a saturated solution of silver carbonate will contain #4.7 * 10^(-5)# moles of dissociated silver carbonate for every #"1 L"# of solution.

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 #color(red)(2)# moles of silver(I) cations and #1# mole of carbonate anions.

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 #25^@"C"#, the saturated solution will contain

#["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.