What is the concentration of Ca2+ ions in a solution of CaI2 that has an I- concentration of 0.520M?

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The dissolution of CaI2 in the solution can be represented as:

`CaI_2 -> Ca^(2+) _(aq) + 2 I^(-) _(aq)`

The mole ratio for this can be said as: 1 mole CaI2; 1 mole Ca2+; 2 moles I-. This can also mean that per one mole of CaI2 dissolved in the solution, 1 mole of calcium ion is formed and 2 moles of iodine ions are formed. There is twice as many iodine ions formed than calcium ions.

**So if we have 0.520 M concentration of I- solution, the concentration for Ca2+ ion is 0.260 M.**

Let 1 L solution:

`0.520 (mol es I^(-))/(L) * (1 L) = 0.520 mol es I^(-)`

`0.520 mol es I^(-) * (1 mol e Ca^(2+))/(2 mol es I^(-)) = 0.260 mol es Ca^(2+)`

Therefore:

`(0.260 mol es Ca^(2+))/1L = 0.260 M`

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