find the molar solubilty of silver chloride in 0.20 M NaCl (aq) with ksp= 1.6x10^10 and iron (II) Hydroxide in

Question

1.0x10^-4 M FeCl2 (aq) Ksp Fe (OH)2 = 1.6X10^-14

Answer 1

First of all the Ksp of AgCl should be 1.6*10^-10 (you forgot the minus sign)

.. .. .. .. .. .. AgCl <=> Ag+ + Cl-
Initial .. .. .. .. .. .. .. .. .. .. .. .. 0.2
Dissolve .. .. x
Produce .. .. .. .. .. .. .. x .. .. ..x
At Equil .. .. .. .. .. .. .. ..x .. 0.2+x

Ksp =[Ag+][Cl-] =x(0.2+x)
Now we ought to solve the quadratic. However we can avoid that by doing the assumption that x << 0.2 and 0.2+x=0.2
Then x0.2=1.6*10^-10 => x= 8*10^-10 M (which of course means that our assumption is valid)

.. .. .. .. .. .. .. Fe(OH)2 <=> Fe+2 + 2OH-
Initial .. .. .. .. .. .. .. .. .. .. .. .. 10^-4
Dissolve .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. ..x .. .. .. 2x
At Equil .. .. .. .. .. .. .. .. .. ..x+10^-4 .. 2x

Ksp =[Fe+2][OH-]^2 =(x+10^-4)*4x^2 =1.6*10^-14
If we do again the assumption x << 10^-4 we get
(4*10^-4) x^2= 1.6*10^-14 =>
x^2 =0.4*10^-10 =>
x= sqrt(0.4*10^-10)= 6.32*10^-6, it is not too much smaller than 10^-4 so maybe we should get a more accurate solution by solving the cubic equation

4x^3+ 4*10^-4 x^2 -1.6*10^-14=0

the only acceptable solution is

x= 6.14 *10^-6 M

Answer 2

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