Question

A solution is prepared by mixing 0.10 L of 0.12 M sodium chloride with 0.23 L of a 0.19 M MgCl2 solution. What volume of a 0.20 M silver nitrate solution is required to precipitate all the Cl− ion in the solution as AgCl?

Answer 1

You need approximately 0.468 L of the 0.20 M silver nitrate solution to precipitate all the Cl- ions in the prepared solution as AgCl.

Calculate the total moles of Cl- ions.

From NaCl: moles of Cl- = concentration (M) * volume (L) = 0.12 M * 0.10 L = 0.012 moles

From MgCl2: since each MgCl2 molecule has 2 Cl- ions, moles of Cl- = 2 * concentration (M) * volume (L) = 2 * 0.18 M * 0.23 L = 0.0816 moles

Total moles of Cl- = 0.012 moles + 0.0816 moles = 0.0936 moles

Determine the moles of AgNO3 needed.

The balanced chemical equation for the precipitation reaction is: AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

The reaction ratio between AgNO3 and Cl- is 1:1, meaning for every 1 mole of Cl-, we need 1 mole of AgNO3.

Therefore, moles of AgNO3 required = moles of Cl- = 0.0936 moles

Calculate the volume of AgNO3 solution.

Volume (L) = moles / concentration (M)

Volume of AgNO3 solution = 0.0936 moles / 0.20 M = 0.468 L