Question

At a certain temperature, K = 1.1×10³ for the reaction

Fe³⁺(aq) + SCN^-(aq) FeSCN²⁺(aq)

Calculate the concentrations of Fe³⁺, SCN^-, and FeSCN²⁺ at equilibrium if 0.020 mole of Fe(NO₃)₃ is added to 1.0 L of 0.10 M KSCN. (Neglect any volume change.)

Short answer

The concentrations of Fe³⁺, SCN^-, and FeSCN²⁺ at equilibrium are ${\text{2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}4}}\text{\hspace{0.17em}M}$,$\text{0.080\hspace{0.17em}M}$ and$\text{0.020\hspace{0.17em}M}$ respectively.

Step-by-step solution

Equillibrium constant

The expression for the equilibrium constant, K of this reaction in terms of concentrations is written below:

$\text{K=}\frac{\left[{\text{FeSCN}}^{\text{2+}}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]\left[{\text{SCN}}^{\text{\hspace{0.17em}–\hspace{0.17em}}}\right]}$ …(1)

The number of moles of${\text{Fe}}^{\text{3+}}$is given as 0.020 mol.

The volume of the container is given as 1.0 L.

Hence, the concentration of${\text{Fe}}^{\text{3+}}$ is,

$\begin{array}{c}\left[{\text{Fe}}^{\text{3+}}\right]\text{=}\frac{{\text{Moles\hspace{0.17em}of\hspace{0.17em}Fe}}^{\text{3+}}}{\text{Volume}}\\ \text{=}\frac{\text{0.020\hspace{0.17em}mol}}{\text{1.0\hspace{0.17em}L}}\\ \text{=0.020\hspace{0.17em}M}\end{array}$

ICE Table

A 0.020 mol of${\text{Fe}}^{\text{3+}}$will react with 0.10 mol of ${\text{SCN}}^{\text{-}}$to produce 0.020 mol of${\text{FeSCN}}^{\text{2+}}$ because the equilibrium constant value is very large.

Let x be the concentration change.

Setup the ICE table:

$\begin{array}{cccccc}& {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)& \text{+}& {\text{SCN}}^{\text{\hspace{0.17em}–\hspace{0.17em}}}\left(\text{aq}\right)& \stackrel{}{\rightleftharpoons }& {\text{FeSCN}}^{\text{2+}}\left(\text{aq}\right)\\ \text{Iinitial}\left(\text{M}\right)& \text{0}& & \text{0.080}& & \text{0.020}\\ \text{Change}\left(\text{M}\right)& \text{+x}& & \text{+x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}\\ \text{Equilibrium}\left(\text{M}\right)& \text{x}& & \text{0.080+x}& & \text{0.020\hspace{0.17em}–\hspace{0.17em}x}\end{array}$

Equilibrium concentration value

Substitute these equilibrium concentration values in terms of x and the equilibrium constant value into equation (1) to calculate the x value as shown below.

${\text{1.1×10}}^{\text{3}}\text{=}\frac{\text{0.020\hspace{0.17em}–\hspace{0.17em}x}}{\left(\text{x}\right)\left(\text{0.080+x}\right)}$

Solve for x.

${\text{x=2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}4}}$

Concentration of all species

The equilibrium concentrations of Fe³⁺, SCN^-, and FeSCN²⁺ can be calculated as follows:

$\begin{array}{c}\left[{\text{Fe}}^{\text{3+}}\right]\text{=x}\\ {\text{=2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}4}}\text{\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{SCN}}^{\text{\hspace{0.17em}–\hspace{0.17em}}}\right]\text{=0.080+x}\\ {\text{=0.080+2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}4}}\\ \text{=0.080\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{FeSCN}}^{\text{2+}}\right]\text{=0.020\hspace{0.17em}–\hspace{0.17em}x}\\ {\text{=0.020\hspace{0.17em}–\hspace{0.17em}2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}4}}\\ \text{=0.020\hspace{0.17em}M}\end{array}$

Therefore, the concentrations of Fe³⁺, SCN^-, and FeSCN²⁺ at equilibrium are , and respectively.