Questions: Suppose 58.1 g of sodium iodide is dissolved in 300 mL of a 0.60 M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Be sure your answer has the correct number of significant digits.

Solution Steps

First, we need to calculate the number of moles of sodium iodide (NaI) dissolved in the solution. The molar mass of NaI is: \[ \text{Molar mass of NaI} = 22.99 \, (\text{Na}) + 126.90 \, (\text{I}) = 149.89 \, \text{g/mol} \]

Using the given mass of NaI: \[ \text{Moles of NaI} = \frac{58.1 \, \text{g}}{149.89 \, \text{g/mol}} = 0.3876 \, \text{mol} \]

Since each formula unit of NaI provides one iodide anion (I\(^-\)), the moles of iodide anion will be the same as the moles of NaI: \[ \text{Moles of Iodide Anion} = 0.3876 \, \text{mol} \]

The final molarity of the iodide anion in the solution can be calculated using the total volume of the solution, which is 300 mL or 0.300 L: \[ \text{Molarity of Iodide Anion} = \frac{\text{Moles of Iodide Anion}}{\text{Volume of Solution in Liters}} = \frac{0.3876 \, \text{mol}}{0.300 \, \text{L}} = 1.2920 \, \text{M} \]

Final Answer