Question

Suppose 1.04 g of sodium iodide is dissolved in 100. mL of a 47.0 m M aqueous solution of silver nitrate.Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't changewhen the sodium iodide is dissolved in it.Round your answer to 3 significant digits.MX 5 ?

Answer 1

6.938 x 10⁻³ M

Procedure

Data

Mass of sodium iodide (NaI) = 0.104 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO₃) = 47 mM = 0.047M

The molecular mass of sodium iodide is 149.89 g/mol.

Step 1

Find the chemical reaction:

NaI + AgNO₃ -> NaNO₃ + AgI

Step 2

Calculate the number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

`\text{ Moles NaI=0.104 g}\frac{1\text{ mol}}{149.89\text{ g}}=6.938*10^(-4)\text{ mol NaI}`

Step 3

For 1 mole AgNO₃ consumed, we need 1 mole NaI to produce 1 mole of AgI and 1 mole NaNO₃

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) + I-(aq)

Step 4

Calculate iodide ions

For 1 mole NaI, we have 1 mole of Na⁺. Therefore we have 6.938 x 10⁻⁴ moles of NaI and 6.938 x 10⁻⁴ moles Na⁺.

Step 5

Calculate molarity of sodium cation

Molarity = moles Na⁺ / volume

`M=\frac{6.938*10^(-4)\text{ mol }}{0.1\text{ L}}=6.938*10^(-3)\text{ M}`