Respuesta :
Answer : The equilibrium concentration of [tex]SCN^-[/tex] in the trial solution is [tex]4.58\times 10^{-8}M[/tex]
Explanation :
First we have to calculate the initial moles of [tex]Fe^{3+}[/tex] and [tex]SCN^-[/tex].
[tex]\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol[/tex]
and,
[tex]\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol[/tex]
The given balanced chemical reaction is,
[tex]Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)[/tex]
Since 1 mole of [tex]Fe^{3+}[/tex] reacts with 1 mole of [tex]SCN^-[/tex] to give 1 mole of [tex]FeSCN^{2+}[/tex]
The limiting reagent is, [tex]SCN^-[/tex]
So, the number of moles of [tex]FeSCN^{2+}[/tex] = 0.0020 mmole
Now we have to calculate the concentration of [tex]FeSCN^{2+}[/tex].
[tex]\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M[/tex]
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]\epsilon[/tex] = molar absorptivity coefficient
[tex]\epsilon[/tex] and l are same for stock solution and dilute solution. So,
[tex]\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}[/tex]
For trial solution:
The equilibrium concentration of [tex]SCN^-[/tex] is,
[tex][SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}][/tex]
[tex][SCN^-]_{initial}[/tex] = 0.00050 M
Now calculate the [tex][FeSCN^{2+}][/tex].
[tex]C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M[/tex]
Now calculate the concentration of [tex]SCN^-[/tex].
[tex][SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}][/tex]
[tex][SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)[/tex]
[tex][SCN^-]_{eqm}=4.58\times 10^{-8}M[/tex]
Therefore, the equilibrium concentration of [tex]SCN^-[/tex] in the trial solution is [tex]4.58\times 10^{-8}M[/tex]
