Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−) - Leaving Cert Chemistry - Question b - 2012

To calculate the equilibrium concentrations, we must first determine the initial concentrations based on the moles given:

• Initial moles of I2 = 0.0800 M in 1 L = 0.0800 M
• Initial moles of I− = 0.2400 M in 1 L = 0.2400 M
• At equilibrium, moles of I3− = 0.0793 M

Using the change in concentration to find equilibrium concentrations:

• For I2, the change is:
  • 0.0800 M - x = 0.0800 M - 0.0793 M = 0.0007 M
• For I−:
  • 0.2400 M - x = 0.2400 M - 0.0793 M = 0.1607 M

Thus the equilibrium concentrations are:

• [I2] = 0.0007 M
• [I−] = 0.1607 M
• [I3−] = 0.0793 M

Now substituting values into the equilibrium constant expression:

$K_c = \frac{0.0793}{0.0007 \times 0.1607} = \frac{704.95}{705}$

So, the equilibrium constant Kc is approximately 704.95.

When a substance that reacts with iodine, such as starch, is added, the equilibrium concentration of triiodide ions (I3−) will decrease.

This occurs because the addition of starch effectively removes some iodine from the solution. According to Le Chatelier's principle, the equilibrium will shift to the left (toward reactants) to restore the concentration of iodine, resulting in a decrease in the concentration of triiodide ions. The system seeks to counteract the change, thus shifting back to re-establish equilibrium.