Question

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

Answer 1

Answer: 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:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$    .....(1)

• For copper (II) iodide:

Molarity of copper (II) iodide solution = 0.272 M

Volume of solution = 49.5 mL

Putting values in equation 1, we get:

$0.272M=\frac{\text{Moles of copper (II) iodide}\times 1000}{49.5}\\\\\text{Moles of copper (II) iodide}=\frac{0.272\times 49.5}{1000}=0.00135mol$

1 mole of copper (II) iodide $(CuI_2)$ produces 1 mole of copper ions $(Cu^{2+})$ and 2 moles of iodide ions $(I^{-})$

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

• For magnesium iodide:

Molarity of magnesium iodide solution = 0.674 M

Volume of solution = 10.6 mL

Putting values in equation 1, we get:

$0.674M=\frac{\text{Moles of magnesium iodide}\times 1000}{10.6}\\\\\text{Moles of magnesium iodide}=\frac{0.674\times 10.6}{1000}=0.00714mol$

1 mole of magnesium iodide $(MgI_2)$ produces 1 mole of magnesium ions $(Mg^{2+})$ and 2 moles of iodide ions $(I^{-})$

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:

$\text{Molarity of }I^{-}\text{ ions}=\frac{0.01698\times 1000}{60.1}\\\\\text{Molarity of }I^{-}\text{ ions}=0.282M$

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