Ferric ions left in the solution at equilibrium is:
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 XV =0.00150*.0700L=0.00105
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' X V'=0.200*0.700L=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
=0.00075m
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
=0.1M
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 =
=
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) =
Learn more about thiocyanate ion, SCN-:
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