Accounting for the change in concentration due to dilution is a critical aspect of this experiment. Calculate the concentration of the I⁻ ions in a solution if 6.0 mL of 0.30 M KI is added to 4.0 mL of 0.45 M NaI and 2.0 mL of water.

The concentration of iodide ion in the resulting solution is 0.225 M

Explanation:

To calculate the molarity of solution, we use the equation:

       .......(1)

Molarity of solution = 0.30 M

Volume of solution = 6.00 mL

Putting values in equation 1, we get:

1 mole of KI produces 1 mole of ions and 1 mole of ions

Moles of iodide ions in KI solution = (1 × 0.0018) = 0.0018 moles

Molarity of solution = 0.45 M

Volume of solution = 2.00 mL

Putting values in equation 1, we get:

1 mole of NaI produces 1 mole of ions and 1 mole of ions

Moles of iodide ions in NaI solution = (1 × 0.0009) = 0.0009 moles

Now, calculating the chloride ions in the solution by using equation 1, we get:

Total moles of iodide ions = [0.0018 + 0.0009] =  0.0027 moles

Total volume of base solution = [6 + 4 + 2] = 12 mL

Putting values in equation 1, we get:

Hence, the concentration of iodide ion in the resulting solution is 0.225 M

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