Answer:
The final molarity of iodide anion is 0.053 M
Explanation:
<u>Step 1</u>: Data given
Mass of sodium iodide (NaI) = 0.795 grams
Volume of the solution = 100 mL = 0.1 L
Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M
The molecular mass of sodium iodide is 149.89 g/mol.
<u>Step 2:</u> The balanced equation
AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)
<u>Step 3: </u>Calculate number of moles of sodium iodide
Moles NaI = mass NaI / Molar mass NaI
Moles NaI = 0.795 grams / 149.89 g/mol
Moles NaI = 0.0053 moles
For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3
The sodium iodide will dissociate as followed:
NaI(aq) → Na+(aq) + I-(aq)
<u>Step 4</u>: Calculate iodide ions
For 1 mole NaI, we have 1 mole of I-
For 0.0053 moles of NaI we'll have 0.0053 moles I-
<u>Step 5:</u> Calculate molarity of iodide ion
Molarity = moles I- / volume
Molarity I- = 0.0053 moles / 0.1 L
Molarity I- = 0.053 M
The final molarity of iodide anion is 0.053 M
