Answer:
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
