Destroyer problem...2016 # 132 GC

[Fancy312]

Hiiii all,

I am really struggling with this destroyer gen chem problem, number 132. If anyone can help I would really appreciate it!

132. The Ksp of Fe(Oh)2 at 25 C is 1.6 x 10 ^-14, What is the solubility of the Fe(OH)2 in 0.025 M of FeCl2?

*** I looked at the solution in the back, using the common ion effect and then solving after the additional source of Fe from FeCl2 BUT I WAS STILL SO CONFUSED ON HOW THEY GOT THE ANSWER. Maybe its just late and im hallucinating???? If anyone can break it down step by step and just help me you would be a lifesaver

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[nossdt]

The first thing you want to do with any solubility problem is to write out your equation for Ksp. In this case we have Fe(OH)2 --> Fe2+ + 2OH- so our Ksp = [Fe2+][OH]^2.

As you noticed, there’s going to be the common ion effect involved here. That’s because Fe(OH)2 is being dissolved in 0.025M of FeCl2. Thus, we already have 0.025 M of Fe2+ ions in our solution before we’ve added ANY Fe(OH)2. Now let’s make our ICE table.

Fe(OH)2 --> Fe2+ + 2OH-

I 0.025 0

C +x +2x

E 0.025 +x 2x

Now we can finally set up our equation and solve for x.

Ksp = 1.6*10^-14 = (0.025+x)(2x)^2.

The trick to this problem is this: since x will always be a REALLY in this kind of solubility problem (0.025 + “a really small number”) we can basically disregard the x in (0.025 +x). ex. 0.025 + 0.0000000000000000000001 = 0.025. This will also make solving for x a LOT easier since it helps us avoid solving a quadratic function. Now we just plug and chug.

1.6* 10^-14 = 0.025(2x)^2.

= (2.5*10^-2)(4x^2)

= (10*10^-2)(x^2)

1.6*10^-13 = x^2

16 * 10^-14 = x^2

X = 4 *10^-7


Hope this answers your question.

 

[Fancy312]

THANKYOUSOMUCH