Question

We prepare a solution by mixing 0.10 L of 0.12 M sodium chloride with 0.23 L of a 0.18 M \(\mathrm{MgCl}_{2}\) solution. What volume of a \(0.20 \mathrm{M}\) silver nitrate solution do we need to precipitate all the \(\mathrm{Cl}^{-}\) ion in the solution as AgCl?

Short answer

The volume of 0.20 M AgNO3 solution required is equal to the total moles of Cl- divided by 0.20 M.

Step-by-step solution

Calculate moles of Cl- from sodium chloride

Multiply the volume of the sodium chloride solution by its molarity to find the moles of NaCl. Each NaCl yields one Cl-. Moles of Cl- from NaCl is given by: moles of NaCl = 0.10 L * 0.12 M.

Calculate moles of Cl- from magnesium chloride

Multiply the volume of the magnesium chloride solution by its molarity to find the total moles of MgCl2. Each MgCl2 yields two Cl-. Moles of Cl- from MgCl2 is given by: moles of MgCl2 = 0.23 L * 0.18 M, total moles of Cl- = moles of MgCl2 * 2.

Calculate total moles of Cl-

Add the moles of Cl- from NaCl to the moles of Cl- from MgCl2 to get the total moles of Cl- in the solution.

Determine Ag+ needed from silver nitrate

Since each mole of AgCl requires one mole of Ag+, the total moles of Ag+ needed is equal to the total moles of Cl-.

Calculate the volume of the 0.20 M AgNO3 solution required

Divide the total moles of Ag+ needed by the molarity of the AgNO3 solution to find the volume required. Volume of AgNO3 = moles of Ag+ / 0.20 M.

Key concepts

Molarity Calculations

When we talk about molarity, we're referring to the concentration of a solution, that is, how much of a substance is dissolved in a specified volume of liquid. Expressed in moles per liter (M), it's a vital concept in stoichiometry as it allows us to understand the relationship between the amount of reactant and the volume of the solution it's contained in.

To calculate molarity, you need to divide the number of moles of the solute by the volume of the solution in liters. For instance, if you have 0.12 moles of sodium chloride (NaCl) dissolved in 0.10 liters of water, the molarity (M) would be calculated as follows:\[M = \frac{0.12 \text{ moles}}{0.10 \text{ L}} = 1.2 \text{ M}\]
Understanding molarity is crucial for precisely mixing chemicals to achieve the desired reaction, such as in this exercise where we are required to calculate the volume of a silver nitrate solution to completely react with the chloride ions present in a mixture.

Mole-to-Mole Ratio

In chemical reactions, the mole-to-mole ratio is the bridge between the reactants and the products. It's the heart of stoichiometry. This ratio comes from the coefficients of each substance in a balanced chemical equation and tells you how many moles of one substance react with or produce another substance.

For instance, in a reaction where one mole of sodium chloride (NaCl) reacts to form one mole of silver chloride (AgCl), the mole-to-mole ratio is 1:1. However, in our current problem where one mole of magnesium chloride (MgCl2) dissociates, it produces two moles of chloride ions (Cl-), giving a mole-to-mole ratio of 1:2. This ratio is critical for determining the precise amount of silver nitrate needed to precipitate the chloride ions.\[\text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^{-}\]

• One mole of MgCl2 yields two moles of Cl- ions.
• Therefore, if you have the total moles of MgCl2, you multiply this number by 2 to get the total moles of Cl-.

By understanding the mole-to-mole ratio, you can seamlessly continue with the stoichiometric calculations necessary for arriving at the solution.

Precipitate Formation

Precipitation is the process where a soluble reactant forms an insoluble product that falls out of the solution as a solid, known as the precipitate. This occurs when two aqueous solutions, each containing a different ion, react to form an insoluble ionic compound.

In this scenario, we're interested in the formation of silver chloride (AgCl). When a silver nitrate (AgNO3) solution is mixed with a solution containing chloride ions (Cl-), a white solid of AgCl precipitates out of the mixture:\[\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} (s)\]

• Each Ag+ ion combines with one Cl- ion to form AgCl.
• For complete precipitation, the amount of Ag+ ions must be equal to the amount of Cl- ions in the mixture.

In our exercise, we find the volume of the silver nitrate solution necessary to provide enough Ag+ ions to completely react with all the chloride ions in the solution. This ensures that all the chloride ions are precipitated and none remain dissolved.