Question

a stock solution is prepared by dissolving 4.68 grams of magnesium iodide in enough water to make 75.0 ml of solution. when a 15.00 ml sample of this stock solution is added to 40.00 ml of water, what is the final concentration of iodide ions?

Answer 1

The final concentration of iodide ions in the solution, after adding a 15.00 ml sample of the stock solution to 40.00 ml of water, is
`\(_\text{I}^- \)` 0.072 M.

Step-by-step explanation:

In order to determine the final concentration of iodide ions, we can follow a step-by-step process. Firstly, let's find the moles of magnesium iodide in the stock solution. The molar mass of magnesium iodide
`(\( \text{MgI}_2 \))` is calculated as follows:

`\[\text{Molar mass of MgI}_2 = (\text{Atomic mass of Mg}) + 2 * (\text{Atomic mass of I}) = 24.31 + 2 * 126.90 \, \text{g/mol} = 277.11 \, \text{g/mol}\]`

Next, we find the moles of magnesium iodide in the stock solution:

`\[\text{Moles of MgI}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4.68 \, \text{g}}{277.11 \, \text{g/mol}} \approx 0.0169 \, \text{mol}\]`

Now, let's calculate the concentration of iodide ions in the stock solution:

`\[\text{Concentration of }_\text{I}^- \text{ ions} = \frac{\text{Moles of }_\text{I}^-}{\text{Volume of solution}} = \frac{0.0169 \, \text{mol}}{0.075 \, \text{L}} \approx 0.225 \, \text{M}\]`

When a 15.00 ml sample is taken and added to 40.00 ml of water, the total volume becomes 55.00 ml. Now, we can find the final concentration:

`\[\text{Final concentration of }_\text{I}^- \text{ ions} = \frac{\text{Initial concentration} * \text{Initial volume}}{\text{Final volume}} = \frac{0.225 \, \text{M} * 15.00 * 10^(-3) \, \text{L}}{55.00 * 10^(-3) \, \text{L}} \approx 0.072 \, \text{M}\]`

Therefore, the final concentration of iodide ions is
`\(_\text{I}^- \)` 0.072 M.