A solution is prepared by mixing 0.10 L of 0.12 M sodium chloride with 0.23 L of a 0.18 M magnesium chloride solution. What volume of a silver nitrate solution (at 0.200 M) is required to precipitate all the Cl– ion in the solution as AgCl (s)?

The volume of silver nitrate solution required is 0.474 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

Molarity of sodium chloride = 0.12 M

Volume of solution = 0.10 L

Putting values in above equation, we get:

1 mole of sodium chloride produces 1 mole of chloride ions and 1 mole of sodium ions

So, moles of chloride ions = 0.012 moles

Molarity of magnesium chloride = 0.18 M

Volume of solution = 0.23 L

Putting values in above equation, we get:

1 mole of magnesium chloride produces 1 mole of magnesium ions and 2 moles of chloride ions

So, moles of chloride ions =

Total moles of chloride ions = [0.012 + 0.0828] = 0.0948 moles

The chemical equation for the reaction of chloride ions and silver nitrate follows:

By Stoichiometry of the reaction:

1 mole of chloride ions reacts with 1 mole of silver nitrate

So, 0.0948 moles of chloride ions will react with = of silver nitrate

Molarity of silver nitrate = 0.200 M

Moles of silver nitrate = 0.0948 moles

Putting values in above equation, we get:

Hence, the volume of silver nitrate solution required is 0.474 L

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