Question

Potassium thiocyanate, KSCN, is often used to detect the presence of Fe3+ ions in solution through the formation of the red Fe(H2O)5SCN2+ (or, more simply, FeSCN2+). What is [Fe3+] when 0.700 L each of 0.00150 M Fe(NO3)3 and 0.200 M KSCN are mixed? Kf of FeSCN2+ = 8.9 × 102. Enter your answer in scientific notation. Report your final answer to two significant figures.

Answer 1

Answer:

Ferric ions left in the solution at equilibrium is $8.0\times 10^{-6} M$.

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

$n=M\times V=0.00150 M\times 0.700 L=0.00105 mol$

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

$n'=M'\times V'=0.200 M\times 0.700 L=0.140 mol$

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

$M_1=\frac{0.00105 mol}{0.700 L+0.700 L}=0.00075 M$

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

$M_2=\frac{0.140 mol}{0.700 L+0.700 L}=0.1 M$

Complex equation:

          $Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}$

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 =$K_f=8.9\times 10^2$

$K_f=\frac{[[Fe(SCN)]^{2+}]}{[[Fe^{3+}]][SCN^{-}]}$

$8.9\times 10^2=\frac{x}{(0.00075 M -x)\times (0.1 M-x)}$

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)= $8.0\times 10^{-6} M$