Asked by Jakob
Approximately 1.5 10-3 g of iron(II) hydroxide, Fe(OH)2(s) dissolves per liter of water at 18°C. Calculate Ksp for Fe(OH)2(s) at this temperature.
This is what I've done:
1.5x10^-3 g Fe(OH)2/L x 1mol Fe(OH)2/89.863g Fe(OH)2 = 1.669x10^-5 M Fe(OH)2
Fe(OH)2 = Fe + 2OH
Ksp = [Fe][OH]^2
Ksp = [1.669x10^-5][1.669x10^-5]^2
Ksp = 4.6x10^-5
I'm told the answer is wrong. Can anyone please tell me where I'm messing up? Thanks!
This is what I've done:
1.5x10^-3 g Fe(OH)2/L x 1mol Fe(OH)2/89.863g Fe(OH)2 = 1.669x10^-5 M Fe(OH)2
Fe(OH)2 = Fe + 2OH
Ksp = [Fe][OH]^2
Ksp = [1.669x10^-5][1.669x10^-5]^2
Ksp = 4.6x10^-5
I'm told the answer is wrong. Can anyone please tell me where I'm messing up? Thanks!
Answers
Answered by
bobpursley
Ksp=[Fe}{OH}^2 = n*(2n)^2 where your n is as you stated. The OH is twice the Fe
Answered by
Jakob
I don't understand... I thought OH is only raised to the 2nd power? Not multiplied by 2 and then raised to the 2nd power?
So the Ksp should be [Fe][2OH]^2?
So the Ksp should be [Fe][2OH]^2?
Answered by
Jakob
Sorry, I understand now... thank you very much for your help!
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