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!
Ksp=[Fe}{OH}^2 = n*(2n)^2 where your n is as you stated. The OH is twice the Fe
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?
Sorry, I understand now... thank you very much for your help!
Well, I hate to break it to you, but it looks like you're "messing up" in a big way. But don't worry, I'm here to help with a little humor!
It seems you forgot to account for the fact that Fe(OH)2 dissociates into Fe and OH ions. But don't fret, chemistry can be a dizzying dance. Now, let's pop on our clown shoes and try this again, shall we?
First, let's go back to your initial calculation:
1.5x10^-3 g Fe(OH)2/L x 1 mol Fe(OH)2/89.863g Fe(OH)2 = 1.669x10^-5 mol Fe(OH)2
Now, since 1 mol of Fe(OH)2 dissociates into 1 mol of Fe and 2 mol of OH, we need to multiply your result by two to get the concentration of OH ions:
1.669x10^-5 mol Fe(OH)2 x 2 mol OH/1 mol Fe(OH)2 = 3.338x10^-5 mol OH
Now that we have the concentration of OH ions, we can plug it into the Ksp expression:
Ksp = [Fe][OH]^2 = [Fe][3.338x10^-5]^2
But hold your red nose, we still need to find the concentration of Fe ions. Since Fe(OH)2 fully dissociates into Fe and OH, the concentration of Fe ions is the same as the concentration of OH ions:
[Fe] = 3.338x10^-5 mol Fe
Now, let's substitute back into the Ksp expression:
Ksp = [Fe][OH]^2 = 3.338x10^-5 mol Fe x (3.338x10^-5 mol OH)^2
Ksp = 3.338x10^-5 mol Fe x 1.113x10^-9 mol^2 OH
And if we multiply those numbers together (I hope you've got your calculator handy!), we get:
Ksp ≈ 3.712x10^-14
So, there you have it! Ksp for Fe(OH)2 at 18°C is approximately 3.712x10^-14. But remember, in the world of chemistry, laughter is always the best solution!
To calculate the Ksp for Fe(OH)2(s) at a given temperature, you need to use the concentration of Fe(OH)2 and consider the stoichiometry of the reaction. Here's a step-by-step explanation of how to approach this problem:
1. Start with the given information: 1.5 x 10^-3 g of Fe(OH)2(s) dissolves per liter of water at 18°C.
2. Convert the mass of Fe(OH)2 to moles:
- 1.5 x 10^-3 g Fe(OH)2 x (1 mol Fe(OH)2 / molar mass of Fe(OH)2)
- The molar mass of Fe(OH)2 is 89.863 g/mol (from the formula)
- This gives you the number of moles of Fe(OH)2 dissolved.
3. Use the volume of the solution (1 L) to determine the concentration:
- Concentration (C) = moles of solute (Fe(OH)2) / volume of solution (L)
- In this case, the volume is given as 1 L, so the concentration is just the number of moles.
4. Use the stoichiometry of the reaction to determine the concentration of Fe and OH ions:
- The balanced equation is Fe(OH)2 -> Fe + 2OH
- This tells you that 1 mole of Fe(OH)2 produces 1 mole of Fe ions and 2 moles of OH ions.
- So, the concentration of Fe ions and OH ions is the same: [Fe] = [OH] = concentration of Fe(OH)2.
5. Substitute the concentration values into the Ksp expression:
- Ksp = [Fe][OH]^2 = (concentration of Fe(OH)2) x (concentration of Fe(OH)2)^2
Using this approach, let's calculate the Ksp again:
1.5 x 10^-3 g Fe(OH)2 / (89.863 g/mol) = 1.669 x 10^-5 mol/L (your calculation is correct)
Ksp = (1.669 x 10^-5 mol/L) x (1.669 x 10^-5 mol/L)^2
Ksp = 1.669 x 10^-5 mol/L x 2.785 x 10^-10 mol^2/L^2 = 4.646 x 10^-15 mol^3/L^3
Therefore, the correct value for Ksp at 18°C is 4.646 x 10^-15 mol^3/L^3.