chemistry questions about solubility?

Question

Calculate the solubility of FeCO3(s) in the presence of 0.010 M in Na2CO3 (Ksp = 3.5 x 10^-11)

Calculate the solubility of FeCO3(s) in the presence of 0.010 M KCN. (Kf(Fe(CN)6 4-) = 7.7 x 10^-36) (Ksp = 3.5 x 10^-11)

Answer 1

1)You are not giving the Ka2 for H2CO3, thus we assume that there is no hydrolysis of CO3(-2) and the [CO3(-2)] coming from Na2CO3 is equal to [Na2CO3]=0.01

.. .. .. .. .. .. FeCO3 <=>Fe(+2) + CO3(-2)
Dissolve .. .. x
produce .. .. .. .. .. .. .. .. .. x .. . .. .. .. . x
But you also have CO3(-2) from Na2CO3 thus [CO3(-2)]=0.01+x

Ksp=[Fe(+2)][CO3(-2)]= x(0.01+x)= 3.5*10^-11

Let's do the approximation that 0.01>> x and 0.01+x=0.01. Then
0.01x=3.5*10^-11 => x=3.5*10^-9 M << 0.01 so our approximation is justified (otherwise you would solve the quadratic and get the exact value)


2) Some of Fe(+2) is sequestered in the complex.
So the solubility of FeCO3 will be s=[FeCO3]= [Fe(+2)]+[Fe(CN)6(-4)], since all Fe (free and in complex) comes from dissolved FeCO3.

.. .. .. .. .. .. Fe(+2) + 6CN(-) <=> Fe(CN)6(-4)
Initial .. .. .. .. s .. .. .. .. 0.01
React .. .. .. . x .. .. .. .. 6x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. x
At equil .. .. s-x .. .. .. ..0.01-6x .. .. ..x

Kf= [Fe(CN)6(-4)] /[Fe(+2)][CN-]^6 =x / [(s-x)*(0.01-6x)^6] =7.7*10^-36

Let's do the approximation that 0.01-6x=0.01
Then x/[(s-x)*10^-12]= 7.7*10^-36 => x/(s-x)= 7.7*10^-48 =>
x=s*7.7*10^-48/(1+7.7*10^-48)

Also Ksp=[Fe(+2)][CO3(-2)] = (s-x)s =3.5*10^-11

Now you would substitute x in the Ksp equation and find s. However 1+7.7*10^-48=1 and thus x=s*7.7*10^-48 which is practically 0.
So s*s=3.5*10^-11 => s=5.9*10^-6 M (that would mean x=4.5*10^-53 which is essentially 0 and our assumption for the simplification of the equation was also justified; if 51 orders of magnitude are not justifiable than waht is !?!)

meaning that the Kf is so small that practically you don't have to consider the formation of the complex and the presence of CN- will practically not affect the solubility of FeCO3.

Answer 2

do your own work and ask your teacher for help