Question

A chemist obtains 500.0 mL of a solution containing an unknown concentration of calcium iodide, CaI 2. He pipets 20 mL of this solution into a 100 mL volumetric flask and dilutes to the mark. He then pipets 10 mL of this diluted solution into a 100 mL volumetric flask and dilutes to the mark. He analyzes some of the solution from the final volumetric flask and finds that the iodide ion concentration is 0.00000574 M. (Note: in solution, calcium iodide breaks apart into one Ca 2+ ion for every two I - ions, so a solution that is 1.0 M in CaI 2 is 2.0 M in I -.) Determine the molar concentration of calcium iodide in the original solution.

Answer 1

Answer:

$\large \boxed{1.44 \times 10^{-5} \text{ mol/L}}$

Explanation:

CaI₂(s) ⟶ Ca²⁺ + 2I⁻

1. Original concentration of I⁻

The chemist diluted 20 mL of the original solution to 100 mL

We can use the dilution formula to calculate the original concentration of I⁻.

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{20 mL}\times c_{1} & = & \text{100 mL} \times 5.74 \times 10^{-6} \text{ mol/L}\\20c_{1}& = & 5.74 \times 10^{-4} \text{ mol/L}\\c_{1}& = & 2.87 \times 10^{-5} \text{ mol/L}\\\end{array}$

2. Original concentration of Ca²⁺

The molar ratio is 1 mol Ca²⁺:2 mol I⁻.

$[\text{Ca}^{2+}] =\dfrac{2.87 \times 10^{-5} \text{ mol I}^{-}}{\text{1 L}} \times \dfrac{\text{1 mol Ca}^{2+} }{\text{2 mol I}^{-}} = \mathbf{1.44 \times 10^{-5}} \textbf{ mol/L}\\\\\text{[Ca ^{2+} ] in the original solution was \large \boxed{\mathbf{1.44 \times 10^{-5}} \textbf{ mol/L}} }$