Ag(s) is added to a solution with the following initial concentrations: [Ag+] = 0.200 M, [Fe2+] = 0.100 M, and [Fe3+] = 0.300 M. The following reversible reaction occurs: Ag+(aq) +... Ag(s) is added to a solution with the following initial concentrations: [Ag+] = 0.200 M, [Fe2+] = 0.100 M, and [Fe3+] = 0.300 M. The following reversible reaction occurs: Ag+(aq) + Fe2+(aq) <=> Ag(s) + Fe3+(aq) with Kc = 2.98. Determine each ion's concentration at equilibrium. Read more

Steps to Solve

1. Set up the ICE table

Here's the ICE table based on the given reaction: $Ag^+{(aq)} + Fe^{2+}{(aq)} \rightleftharpoons Ag_{(s)} + Fe^{3+}_{(aq)}$

2. Write the expression for $K_C$

The equilibrium constant $K_C$ is given by:

$K_C = \frac{[Fe^{3+}]}{[Ag^+][Fe^{2+}]}$

Note that we don't include $Ag(s)$ in the $K_C$ expression because it is a solid.

3. Substitute the equilibrium concentrations into the $K_C$ expression

We are given that $K_C = 2.98$. Substituting the equilibrium concentrations from the ICE table into the $K_C$ expression gives

$2.98 = \frac{0.300 + x}{(0.200 - x)(0.100 - x)}$

4. Solve for x

Expanding the equation:

$2.98 = \frac{0.300 + x}{0.02 - 0.300x + x^2}$

$2.98(0.02 - 0.300x + x^2) = 0.300 + x$

$0.0596 - 0.894x + 2.98x^2 = 0.300 + x$

Rearrange to form a quadratic equation:

$2.98x^2 - 1.894x - 0.2404 = 0$

Use the quadratic formula to solve for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{1.894 \pm \sqrt{(-1.894)^2 - 4(2.98)(-0.2404)}}{2(2.98)}$

$x = \frac{1.894 \pm \sqrt{3.587236 + 2.865952}}{5.96}$

$x = \frac{1.894 \pm \sqrt{6.453188}}{5.96}$

$x = \frac{1.894 \pm 2.540}{5.96}$

We have two possible solutions for $x$:

$x_1 = \frac{1.894 + 2.540}{5.96} = \frac{4.434}{5.96} \approx 0.744$

$x_2 = \frac{1.894 - 2.540}{5.96} = \frac{-0.646}{5.96} \approx -0.108$

Since concentrations cannot be negative, we must have $x = -0.108$ rejected. However, $x$ also can't be greater than 0.1 because that would provide a negative concentration for $Fe^2+$. Therefore, the result $x_1 \approx 0.744$ must be rejected. A likely sign that something is wrong with an earlier step. Let's re-examine the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{1.894 \pm \sqrt{(-1.894)^2 - 4(2.98)(-0.2404)}}{2(2.98)}$

$x = \frac{1.894 \pm \sqrt{3.587236 + 2.865952}}{5.96}$

$x = \frac{1.894 \pm \sqrt{6.453188}}{5.96}$

$x = \frac{1.894 \pm 2.540}{5.96}$

$x_1 = (1.894 + 2.54) / 5.96 = 4.434 / 5.96 = 0.744 \leftarrow$ rejected

$x_2 = (1.894 - 2.54) / 5.96 = -0.646 / 5.96 = -0.108 \leftarrow$ rejected

Hmm. Is there a possibility that the reaction flows in reverse? Let us assume that concentrations of $Ag^+$ and $Fe^{2+}$ increases, while the concentration of $Fe^{3+}$ decreases.

$2.98 = \frac{0.300 - x}{(0.200 + x)(0.100 + x)}$

$2.98 = \frac{0.300 - x}{0.02 + 0.3x + x^2}$

$2.98(0.02 + 0.3x + x^2) = 0.300 - x$

$0.0596 + 0.894x + 2.98x^2 = 0.300 - x$

$2.98x^2 + 1.894x - 0.2404 = 0$

$x = \frac{-1.894 \pm \sqrt{(1.894)^2 - 4(2.98)(-0.2404)}}{2(2.98)}$

$x = \frac{-1.894 \pm \sqrt{3.587236 + 2.865952}}{5.96}$

$x = \frac{-1.894 \pm \sqrt{6.453188}}{5.96}$

$x = \frac{-1.894 \pm 2.540}{5.96}$

$x_1 = (-1.894 + 2.54) / 5.96 = 0.646 / 5.96 = 0.108$

$x_2 = (-1.894 - 2.54) / 5.96 = -4.434 / 5.96 = -0.744 \leftarrow$ rejected

Therefore $x = 0.108$ and solving

$[Ag^+] = 0.200 + 0.108 = 0.308 M$

$[Fe^{2+}] = 0.100 + 0.108 = 0.208 M$

$[Fe^{3+}] = 0.300 - 0.108 = 0.192 M$