How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

Chemistry

1 answer:

KiRa [2.9K]3 months ago

4 0

Response:

To reach the answer, 465.6 mg of MgI₂ is required.

Detailed Explanation:

We need to establish the moles of ion I⁻ in the resulting solution.

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol.

In the initial solution, there was 0.087 M KI, which we can similarly convert into moles, yielding 0.02242 mol.

This indicates we require an additional amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. Since each molecule of MgI₂ produces two I⁻ ions, we divide 0.00335 by 2 to determine the moles of MgI₂, giving us 0.001675 mol.

Consequently, the quantity of MgI₂ to be added is:

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

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