Suppose 2.07g of sodium iodide is dissolved in 50.mL of a 0.30M aqueous solution of potassium carbonate. 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. Round your answer to 2 significant digits.

0.28M is molarity of iodide anion in the solution.

Explanation:

Molarity is defined as the ratio between moles of solute (In this case, iodine ion), per liter of solution (The total volume of the solution is 50.0mL = 0.0500L).

Thus, we need to convert mass of sodium iodide to moles using its molar mass (Molar mass NaI: 149.89g/mol). Moles of NaI = Moles of I⁻:

Moles NaI = Moles iodide ion:

2.07g * (1mol / 149.89g) = 0.01381moles NaI = Moles I⁻

Molarity is:

0.01381moles I⁻ / 0.0500L =

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it's filtration, but this is an assumption so my apologies if i got this