[NewtoAtoms]

Would someone please tell me if my leg work and logic behind this question is correct?

Q: 2.20g of solid NaOH are added to 250 mL of 0.10 M FeCl2 solution.  Given the Ksp for Fe(OH)2 is 1.6 x 10-14 calculate:

a.  the weight of Fe(OH)2 formed
b.  the molar concentration of Fe2+ in the final solution.

I have done all the leg work, which makes perfect sense to me, unfortunately I am extremely worried that I am wrong.
Would you chemists please review my work and tell me if I am on the correct path?

a. weight of Fe(OH)₂ formed

g FeCl₂ ----> mol FeCl₂ ----> mol fe(OH)₂ ----> g Fe(OH)₂

2.20g NaOH x 1mol/41.01 gNaOH x 1 mol Fe(OH)2/1 mol NaOH x 73.87 g/1 mol Fe(OH)2

= 3.99 g FeOH₂- is what I would propose is the weight of Fe(OH)² formed.  I calculated this using the stoichiometric ratios.

b. the molar concentraiton of Fe2+ in the final solution

            Fe(OH)2  <----------> Fe²⁺        2(OH^-)
INITIAL                                   O               O
CHANGE                                  +s             +2s
EQUIL                                       S              2S

Ksp = [Fe][OH]²
Ksp = [ S][2S]²
Ksp = [ S] [4S²]
1.6 x 10^-14 = 4s3  (divide both sides by four, then cube root right side)
1.59 x 10^-5 = s

Therefore the molar concentration of Fe²⁺ in the final solution would be 1.59 x 10^-5

Any help, direction and time would be deeply appreciated!!

[NewtoAtoms]

I am sorry I am not sure why the end 1/2 of the question has a line through it...

[AWK]

Check this number

41.01