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

Ferric ions left in the solution at equilibrium is
`8.0* 10^(-6) M`.

Step-by-step 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* V=0.00150 M* 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'* V'=0.200 M* 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=(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=(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* 10^2`

`K_f=([[Fe(SCN)]^(2+)])/([[Fe^(3+)]][SCN^(-)])`

`8.9* 10^2=(x)/((0.00075 M -x)* (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* 10^(-6) M`