If 67.1 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.235 g of precipitate, what is the molarity of silver ion in the original solution?

0.098 M

Explanation:

The equation for the reactions is given by;

AgNO₃ (aq) + KCl (aq) → AgCl (s) + KNO₃ (aq)

Volume of silver nitrate is 67.1 mL Mass of the precipitate, AgCl is 0.235 g

We are required to calculate the molarity of silver ion in the original solution;

Step 1: Calculate the moles of AgCl

We know that, Moles = Mass : Molar mass

Molar mass of AgCl = 143.32 g/mol

Therefore;

Moles of AgCl = 0.235 g : 143.32 g/mol

= 0.00164 moles

Step 2: Moles of silver nitrate

From the equation, 1 mole of silver nitrate reacts to yield 1 mole of AgCl

Therefore, Moles of silver nitrate = moles of AgCl

Thus, moles of silver nitrate = 0.00164 moles

Step 3: Moles of silver ions

Moles of silver nitrate = 0.00164 moles

But;

AgNO₃ (aq) → Ag⁺ (aq) + NO₃⁻ (aq)

Therefore, moles of Ag⁺ is equal to the moles of AgNO₃

Moles of Ag⁺ = 0.00164 moles

Step 4: Molarity of silver ions in silver nitrate

Molarity = Moles : Volume

Therefore;

Molarity of Ag⁺ = 0.00164 moles : 0.0671 L

= 0.0982 Moles/L

= 0.098 M

Therefore, the molarity of silver ions in silver nitrate is 0.098 M