Equilibrium ❮ 239❯ Answer and Explanation
(a) Kd = [Cd^2 +][I–]^4 /[CdI 4 2–]
Give yourself 1 point for this expression.
Using the following table:
CdI 4 2–(aq) Cd2+(aq) I–(aq)
Initial 0.100 M 0 0
Change +x +x + 4 x
Equilibrium 0.100 + x x 4 x
The value of [Cd^2 +] is given (= 0.013), and this is x. This changes the last line of
the table to
CdI 42 - (aq) Cd^2 +(aq) I-(aq)
Equilibrium 0.100 + x = 0.087 x = 0.013 4 x = 0.052
Entering these values into the Kd expression gives (0.013) (0.052)^4 /(0.087) = 1.1 × 10 -^6.
Give yourself 1 point for this answer. You can also get 1 point if you correctly put your
values into the wrong equation.
(b) The table in part a changes to the following:CdI 42 - (aq) Cd^2 +(aq) I-(aq)
Initial 0.100 M 0 0.400
Change +x +x + 4 x
Equilibrium 0.100 + x x 0.400 + 4 xKd = [Cd^2 +][I-]^4 /[CdI 42 - ] = 1.1 × 10 -^6 = (x)(0.400 + 4 x)^4 /(0.100 + x)
= (x)(0.400)^4 /(0.100)
x = 4.3 × 10 –6 M = [Cd^2 +]
Give yourself 1 point for the correct setup and 1 point for the correct answer. If you
got the wrong value for K in part a, you can still get one or both points for using it
correctly.
(c) The equilibrium is Cd(OH) 2 ^ Cd^2 + (aq) + 2 OH- (aq)
The dilution reduces both the Cd^2 + and OH- concentration by a factor of 2. This gives
[Cd^2 +] = 4.3 × 10 -^6 /2 = 2.2 × 10 –6 and [OH-] = 2.0 × 10 -^5 /2 = 1.0 × 10 -^5
The reaction quotient is Q = [Cd^2 +] [OH-]^2 = (2.2 × 10 -^6 ) (1.0 × 10 -^5 )^2 = 2.2 × 10 -^16.
This value is less than the Ksp, so no precipitate will form.
Give yourself 1 point for a correct calculation and 1 point for the correct conclusion.
If you correctly use a wrong [Cd^2 +] calculated in part b, you can still get both points.