Suppose 20.8 g of sodium iodide is dissolved in 250. mL of a 0.70 M aqueous solution of silver nitrate. Calculate the final molarity of ALL IONS in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 3 significant digits.

[Ag+] = [NO3-] = 0.700M

0.555M = [Na+] = [I-]

Explanation:

To solve this question we need to find the moles of sodium iodide, NaI, using its molar mass -. With the moles and the volume we can find the molarity of Na+ and I-. The molarity of the ions of silver nitrate, AgNO3 doesn't change because we are assuming the volume doesn't change:

Molarity Ag⁺ = Molarity NO₃⁻ = 0.700M

Moles NaI -Molar mass: 149.89g/mol-

20.8g NaI * (1mol/149.89g) = 0.0139 moles NaI

Molarity:

0.0139 moles NaI / 0.250L = 0.555M = [Na+] = [I-]

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