Respuesta :
Answer:
Ferric ions left in the solution at equilibrium is [tex]8.0\times 10^{-6} M[/tex].
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
[tex]n=M\times V=0.00150 M\times 0.700 L=0.00105 mol[/tex]
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
[tex]n'=M'\times V'=0.200 M\times 0.700 L=0.140 mol[/tex]
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
[tex]M_1=\frac{0.00105 mol}{0.700 L+0.700 L}=0.00075 M[/tex]
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
[tex]M_2=\frac{0.140 mol}{0.700 L+0.700 L}=0.1 M[/tex]
Complex equation:
[tex]Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}[/tex]
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 =[tex]K_f=8.9\times 10^2[/tex]
[tex]K_f=\frac{[[Fe(SCN)]^{2+}]}{[[Fe^{3+}]][SCN^{-}]}[/tex]
[tex]8.9\times 10^2=\frac{x}{(0.00075 M -x)\times (0.1 M-x)}[/tex]
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)= [tex]8.0\times 10^{-6} M[/tex]
