In the laboratory a student combines 49.5 mL of a 0.272 M copper(II) iodide solution with 10.6 mL of a 0.674 M magnesium iodide solution. What is the final concentration of iodide anion ? M

The final concentration of iodide anion in the solution is 0.282 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

   .....(1)

Molarity of copper (II) iodide solution = 0.272 M

Volume of solution = 49.5 mL

Putting values in equation 1, we get:

1 mole of copper (II) iodide produces 1 mole of copper ions and 2 moles of iodide ions

Moles of iodide ions = (2 × 0.00135) = 0.0027 moles

Molarity of magnesium iodide solution = 0.674 M

Volume of solution = 10.6 mL

Putting values in equation 1, we get:

1 mole of magnesium iodide produces 1 mole of magnesium ions and 2 moles of iodide ions

Moles of iodide ions = (2 × 0.0071) = 0.01428 moles

Now, calculating the concentration of iodide anion in the solution by using equation 1, we get:

Total moles of iodide ions = [0.0027 + 0.01428] = 0.01698 moles

Total volume of the solution = [49.5 + 10.6] = 60.1 mL

Putting values in equation 1, we get:

Hence, the final concentration of iodide anion in the solution is 0.282 M.

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