Iron (III) ions form a complex with thiocyanate according to the equation
Fe^3+(aq) + 6 SCN-(aq) ⇄ Fe(SCN)6 ^3-(aq) Kf = 3.20x10^3
If 0.001 moles of Fe(NO3)3 and 0.2 moles of NaSCN are dissolved in enough water to create a 1.00L of solution, what's the concentration of Fe(SCN)6 3- (in M)?
I did:
..................Fe^3+(aq) + 6 SCN-(aq) ⇄ Fe(SCN)6 ^3-(aq)
Initial.........0.001................0.2....................0
change ... +x......................+6x.................-x
equilib......0.001+x~0.001...0.2+6x~ 0.2....-x
Kf = [Fe (SCN)6 ^3-]/ [Fe^3+] [SCN-]^6
3.20x10^3 = (-x)/ ( (0.001)(0.2)^6 )
I got x = -2.1 x 10^-4 M, but it's wrong.
First, your answer of a negative concentration you KNOW can't be right. When have you ever had a solution that had less than nothing as a concentration. Doesn't make sense.
Looking at the equation, you can see that the limiting reagent (LR) in this case is Fe^3+. Therefore, all of the Fe^3+ will be used because the Kf is so large. I would answer that [Fe(SCN)6]^3- = 0.001 M because all of the initial Fe^3+ has formed the complex and by Le Chatelier's Principle the excess SCN^- helps drive the reaction to the right.
Also, I should point out that your ICE chart is going the wrong way. If you have zero Fe(SCN)6^3- it can't be forming the reactants.
Oh no, I'm sorry to hear that your answer is wrong. I guess the chemistry equation couldn't resist clowning around with you. Let's try to solve this together and see if we can find the correct answer.
So, we have the equation Fe^3+(aq) + 6 SCN^-(aq) ⇄ Fe(SCN)6^3-(aq), with the given equilibrium constant Kf = 3.20x10^3.
Let's assume that the equilibrium concentration of Fe(SCN)6^3- is x M. Since Fe(NO3)3 is a strong electrolyte, we can assume that the concentration of Fe^3+ ions is negligible compared to the initial 0.001 moles. Therefore, we can write:
[Fe^3+] = 0.001 M
Now, the concentration of SCN- ions is given as 0.2 moles, but we have to remember that it is 6 times the concentration of SCN- ions in the complex, because of the stoichiometry of the equation. So:
[SCN-] = (0.2 moles / 6 L) = 0.0333 M
Now, we can substitute these values in the equilibrium constant expression:
3.20x10^3 = x / (0.001 M * (0.0333 M)^6)
Solving for x, we find:
x = 3.20x10^3 * 0.001 M * (0.0333 M)^6
And that should give you the correct concentration of Fe(SCN)6^3-. Sorry for the confusion earlier, I hope this helps!
To find the concentration of Fe(SCN)6^3- in the solution, we need to use the given equilibrium constant (Kf) and the initial concentrations of Fe^3+ and SCN-.
Let's start by calculating the initial concentrations of Fe^3+ and SCN-:
Fe(NO3)3 completely dissociates in water, so the initial concentration of Fe^3+ is 0.001 moles / 1 L = 0.001 M.
NaSCN also completely dissociates in water, so the initial concentration of SCN- is 0.2 moles / 1 L = 0.2 M.
Now let's set up the equation using the equilibrium constant (Kf) and the given equation:
Kf = [Fe(SCN)6^3-] / [Fe^3+][SCN-]^6
Substituting the given values:
3.20x10^3 = [Fe(SCN)6^3-] / (0.001)(0.2)^6
Simplifying, we get:
3.20x10^3 = [Fe(SCN)6^3-] / (0.001)(0.000064)
Cross-multiplying and solving for [Fe(SCN)6^3-], we find:
[Fe(SCN)6^3-] = 3.20x10^3 * (0.001)(0.000064)
[Fe(SCN)6^3-] = 0.02048
Therefore, the concentration of Fe(SCN)6^3- in the solution is 0.02048 M.
To find the concentration of Fe(SCN)6 3- (in M), we need to solve for x in the equation:
Kf = [Fe (SCN)6 ^3-] / [Fe^3+] [SCN-]^6
Given that Kf = 3.20x10^3, [Fe^3+] = 0.001 M, and [SCN-] = 0.2 M, we can substitute these values into the equation:
3.20x10^3 = (-x) / (0.001 * (0.2)^6)
Let's proceed step by step to find the correct value of x:
1. Calculate the value of [SCN-]^6:
[SCN-]^6 = (0.2)^6 = 0.000064
2. Substitute the values into the equation:
3.20x10^3 = (-x) / (0.001 * 0.000064)
3. Simplify the equation:
3.20x10^3 = (-x) / 6.4x10^-8
4. Cross multiply the equation:
(-x) = 3.20x10^3 * 6.4x10^-8
5. Calculate the right side of the equation:
(-x) = 2.048x10^-4
6. Solve for x:
x = -2.048x10^-4
Therefore, the concentration of Fe(SCN)6 3- is -2.048x10^-4 M.