Calculate the solubility of iron(II) carbonate at 25 degrees Celsius. The Ksp of FeCO3(s) is 3.5 x 10^-11 at 25 degrees Celsius:
Heres what I have done so far:
1)The balacned equation for this reaction is Fe2CO3 --> 2Fe+ + CO3-
2)KsP = [Fe+][CO3-]
3) Fe2CO3 --> 2Fe+ + CO3-
I NA 0 0
C NA +x +x
E NA x x
4)Ksp = x(x)
3.2 x 10^-11 = x^2
x =
What does x equal?
x equals whatever you let it equal. Usually, in solubility problems, we let x = solubility of FeCO3, then Fe is X and CO3^= = x.
In your problem x = sqrt 3.2 x 10^-11 (should that be 3.5 x 10^-11) and that is (FeCO3). By the way, I don't know how you ended up with equation 3 of Fe2CO3. iron(II) carbonate is FeCO3 as you have it in equation 2 and not in 1 and 3.
I got 5.7 x 10 ^ - 6 why is this incorrect:(
5.7 x 10^-6 is the square root of 3.2 x 10^-11 BUT your problem states at the beginning that Ksp = 3.5 (not 3.2) x 10^-11. As for the equations, iron(II) carbonate is FeCO3 so
FeCO3 ==> Fe^+2 + CO3^=
Ksp + (Fe^+2)(CO3^=)
abd tgise equations where you show Fe2CO3 etc are not right. I get sqrt 3.5 x 10^-11= 5.92 x 10^-6 which I suppose should be rounded to 5.9 x 10^-6
To calculate the solubility of iron(II) carbonate (FeCO3) at 25 degrees Celsius, you need to solve the equation for x that you have set up in step 4.
The equation you have written is correct: 3.5 x 10^-11 = x^2.
To solve for x, take the square root of both sides:
√(3.5 x 10^-11) = √(x^2)
This gives you:
x = √(3.5 x 10^-11)
Using your calculator, evaluate the square root:
x ≈ 5.92 x 10^-6
Therefore, the solubility of iron(II) carbonate at 25 degrees Celsius is approximately 5.92 x 10^-6 moles per liter.