Question

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?

Answer 1

0.098 M

Step-by-step 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