[Kayla927]

Please help... I have been working on this for a long time and just can't see a clear path as to what I am being asked to find!

2.20g of solid NaOH are added to 250mL of a .01M FeCl2 solution. Given ksp for Fe(OH)2 is 1.6 x 10^ -14 calculate the following
A. The mass of Fe( OH)2 formed
B. The molar concentration of Fe in the final solution

Help would be very appreciated!!

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

[Kayla927]

You have to show your attempts at solving the question to receive help. This is a forum policy.

Sorry, first time:
To solve for
A) I used ksp= (fe)(oh)^2
1.6 x 10^-14 = (x) (2x)^2
X= 1.6x10^-5
1.6x10^-5 x 89.87 fe(oh)2 = 1.426x10^-3 x .025=
3.6x10^-5g of Fe(oh)2 dissolved
Now I know this is the amount of (fe) and (oh) dissolved in the solution and not the amount of fe(oh)2 formed so I don't really know where to go from here

B) fe = .025L x .10M/1L= .0025 M of fe
 

[Borek]

ksp= (fe)(oh)^2

And what is the concentration of OH^-?

[Kayla927]

ksp= (fe)(oh)^2

And what is the concentration of OH^-?

OH= 2.2gx 1mol/40.00gNaOHx 2mol/1molx .025x1L= 2.75x10^-3M
Q= (2.5x10^-3)(2.75x10^-3)^2 = 1.8x10^-8
1.8x10^-8x 89.87= 1.6x10^-6g of Fe(OH)2 formed
Is this correct?

[Kayla927]

Also, so to answer the final concentration of Fe
Fe =ksp/2.75x10^-3
Or Fe=Q/2.75x10^-3

[Borek]

OH= 2.2gx 1mol/40.00gNaOHx 2mol/1molx .025x1L= 2.75x10^-3M

While you are right abut starting with 2.2 g of NaOH to calculate OH^- concentration I have no idea what you are doing here, and the final result is wrong.

Q= (2.5x10^-3)(2.75x10^-3)^2 = 1.8x10^-8
1.8x10^-8x 89.87= 1.6x10^-6g of Fe(OH)2 formed
Is this correct?

Sorry to be blunt, but I don't plan to spend my evening guessing which number means what. I have no idea what you are doing.

Hint: start with finding the answer to B, and think if the final concentration of Fe²⁺ is low enough for the precipitation to be (for practical purposes) complete.