The molar solubility of silver chromate at a certain temperature is #2 xx 10^(-4) "mol"cdot"dm"^(-3)#. What will it become in a solution containing #"0.05 M"# #"AgNO"_3# at that same temperature?
1 Answer
Here's what I got.
Explanation:
Silver chromate,
#"Ag"_ color(red)(2)"CrO"_ (4(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "CrO"_(4(aq))^(2-)#
By definition, the solubility product constant,
#K_(sp) = ["Ag"^(+)]^color(red)(2) * ["CrO"_4^(2-)]#
Now, the problem provides you with the molar solubility of silver chromate at a given temperature, which essentially tells you how many moles of silver chromate can be dissolved in water before the resulting solution becomes saturated.
In this case, a molar solubility of
To find the value of the solubility product constant, look at the mole ratios that exist between silver chromate and the dissolved ions, i.e. one mole of silver chromate produces two moles of silver cation and one mole of chromate anions.
If you can only dissolve
#["Ag"^(+)] = color(red)(2) xx 2 * 10^(-4)"M" = 4 * 10^(-4)"M"#
#["CrO"_4^(2-)] = 1 xx 2 * 10^(-4)"M" = 2 * 10^(-4)"M"#
This means that you have
#K_(sp) = (4 * 10^(-4))^color(red)(2) "M"^color(red)(2) * 2 * 10^(-4)"M"#
#K_(sp) = color(green)(barul|stackrel(" ")(" "3.2 * 10^(-11)"mol"^3"dm"^(-9)" ")|)#
To find the molar solubility of silver chromate in a solution that contains silver nitrate,
#"AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO"_(3(aq))^(-)#
Here one mole of silver nitrate produces one mole of silver cations,
#["Ag"^(+)] = "0.05 M"#
Keep in mind that the volume of the solution doesn't matter here because you're looking for concentration, not number of moles. In this regard, the volume could have been given as
Use an ICE table to find the new solubility
#"Ag"_ color(red)(2)"CrO"_ (4(s)) " "rightleftharpoons" " color(red)(2)"Ag"_ ((aq))^(+) " "+" " "CrO"_(4(aq))^(2-)#
This time, you will have
#K_(sp) = (color(red)(2)s + 0.05)^color(red)(2) * s#
#3.2 * 10^(-11) = (4s^2 +0.2s + 0.0025) * s#
Rearrange to get
#4s^3 +0.2s^2 + 0.0025s - 3.2 * 10^(-11)= 0#
This cubic equation will produce one real value
#s = 1.3 * 10^(-8)#
Therefore, the molar solubility of silver chromate in a solution that contains
#s = color(green)(barul|stackrel(" ")(" "1.3 * 10^(-8)"M"" ")|)#
I'll leave the answers rounded to two sig figs.
Notice that the solubility of silver chromate decreases significantly in the presence of the excess silver cations. We saw
This happens because the solubility equilibrium, which is governed by Le Chatelier's Principle, will lie further to the left in the presence of excess silver cations; hence, more
You'll see this referred to as the common-ion effect.