Answer:
Ferric ions left in the solution at equilibrium is
.
Explanation:
Moles of ferric nitrate in 0.700 L = n
Volume of the solution = V = 0.700 L
Molarity of ferric nitrate = M = 0.00150 M

Moles of potassium thiocyanate in 0.700 L = n'
Volume of the potassium thiocyanate solution = V' = 0.700 L
Molarity of potassium thiocyanate = M' = 0.200 M

Molarity of ferric ions after mixing :
1 mol of ferric nitrate gives 1 mol of ferric ions.Then 0.00105 mol ferric nitrate will :
Moles of ferric ions = 0.00105 mol

Molarity of thiocyanate ions after mixing :
1 mol of potassium thiocyanate gives 1 mol of thiocyanate ions.Then 0.140 mol potassium thiocyanate will give:
Moles of thiocyanate ions = 0.140 mol

Complex equation:
![Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%2BSCN%5E-%5Crightleftharpoons%20%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
0.00075 M 0.1 M 0
At equilibrium:
(0.00075 M -x) (0.1 M-x) x
The formation constant of the given complex =
![K_f=\frac{[[Fe(SCN)]^{2+}]}{[[Fe^{3+}]][SCN^{-}]}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%5B%5BFe%28SCN%29%5D%5E%7B2%2B%7D%5D%7D%7B%5B%5BFe%5E%7B3%2B%7D%5D%5D%5BSCN%5E%7B-%7D%5D%7D)

Solving for x:
x = 0.000742 M
Ferric ions left in the solution at equilibrium :
= (0.00075 M -x) = (0.00075 M - 0.000742 M)= 