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2 years ago

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Suppose 7.05 g of zinc iodide is dissolved in 150. mL of a 0.20M aqueous solution of potassium carbonate. Calculate the final mo

larity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the zinc iodide is dissolved in it. Round your answer to 3 significant digits.

1 answer:

sasho [114]2 years ago

4 0

Answer:

$M=0.294M$

Explanation:

Hello!

In this case, since the molarity of the iodide ion is computed by dividing the moles of iodide ions by the volume of solution in liters (0.150 L), we first need to compute the moles of zinc iodide (ZnI2) by using its molar mass and then using a mole ratio of 1 mol of zinc iodide to 2 mol of iodide ion:

$n_{I^-}=7.05gZnI_2*\frac{1molZnI_2}{319.22 gZnI_2}*\frac{2molI^-}{1molZnI_2}=0.0442molI^-$

Therefore, the resulting molarity turns out:

$M=\frac{0.0442molI^-}{0.150L} \\\\M=0.294M$

Best regards!

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