Answer:
Concentration of [tex][Fe^{3+}][/tex] at equilibrium is 7.394 M.
Explanation:
Moles of ferric nitrate = 0.026 mol
Concentration of ferric nitrate = [tex]\frac{0.026 mol}{2.164\times 10^{-3} L}=12.015 M[/tex]
1 mole of ferric nitrate gives 1 mole of ferric ions, then 12.015 M of ferric nitrate solution will have:
[Fe^{3+}]=12.015 M[/tex]
Moles of [tex]Fe(SCN)]^{2+}[/tex] at equilbrium = 0.01 mol
Concentration of [tex]Fe(SCN)]^{2+}[/tex] at equilbrium =
[tex][Fe(SCN)]^{2+}=\frac{0.01 mol}{2.164\times 10^{-3} L}=4.621 M[/tex]
[tex]Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}[/tex]
Initially
12.015 M 0
At equilibrium
(12.015 - 4.621 )M 4.621 M
Concentration of [tex][Fe^{3+}][/tex] at equilibrium
= (12.015 - 4.621 )M = 7.394 M
