Question

A common method for determining how much chloride ion is present in a sample is to precipitate the chloride from an aqueous solution of the sample with silver nitrate solution and then to weigh the silver chloride that results. The balanced net ionic reaction is $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s) $$ Suppose a 5.45 -g sample of pure sodium chloride is dissolved in water and is then treated with a solution containing \(1.15 \mathrm{~g}\) of silver nitrate. Will this quantity of silver nitrate be capable of precipitating all the chloride ion from the sodium chloride sample?

Short answer

The given amount of silver nitrate (1.15 g) will not be enough to precipitate all the chloride ions from the sodium chloride sample (5.45 g). This is because the available moles of silver nitrate (0.00677 mol) are less than the required moles (0.0933 mol) for a complete precipitation based on the stoichiometry of the balanced net ionic reaction.

Step-by-step solution

Calculate moles of sodium chloride (NaCl) in the sample

To find the moles of sodium chloride, we need to divide the mass of the sample by its molar mass. The molar mass of NaCl is 58.44 g/mol (22.99 g/mol for Na and 35.45 g/mol for Cl). Moles of NaCl = (mass of NaCl) / (molar mass of NaCl) Moles of NaCl = (5.45 g) / (58.44 g/mol) = 0.0933 mol

Calculate moles of silver nitrate (AgNO3) in the sample

Similarly, calculate the moles of silver nitrate by dividing the mass of the sample by its molar mass. The molar mass of AgNO3 is 169.87 g/mol (107.87 g/mol for Ag, 14.01 g/mol for N and 16.00 g/mol * 3 for O). Moles of AgNO3 = (mass of AgNO3) / (molar mass of AgNO3) Moles of AgNO3 = (1.15 g) / (169.87 g/mol) = 0.00677 mol

Use stoichiometry to calculate the required amount of silver nitrate

According to the balanced net ionic reaction, there is a 1:1 ratio of Ag+ ions and Cl- ions. This implies that 1 mole of AgNO3 is required to precipitate 1 mole of chloride ions. Moles of AgNO3 required = Moles of Cl- ions Moles of AgNO3 required = 0.0933 mol (from Step 1)

Compare the required amount with the provided amount

Now, we need to compare the amount of silver nitrate required to precipitate all the chloride ions with the given amount of silver nitrate. Moles of AgNO3 required = 0.0933 mol Moles of AgNO3 available = 0.00677 mol Since the available moles of AgNO3 are less than the required moles, the given amount of silver nitrate will not be enough to precipitate all the chloride ions from the sodium chloride sample.

Key concepts

Molar Mass Calculation

Understanding molar mass is critical for solving many problems in chemistry, including stoichiometry exercises. It refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The mole is a fundamental concept in chemistry that represents Avogadro's number of particles, which is approximately 6.022 x 10²³ particles.

When calculating molar mass, we add up the atomic masses of all the atoms in a molecule. For example, sodium chloride (NaCl) has a molar mass of 58.44 g/mol. This is found by summing the atomic mass of sodium (Na), 22.99 g/mol, with that of chlorine (Cl), which is 35.45 g/mol.

To find the moles of a substance, divide the given mass by the molar mass:\[\begin{equation}\text{Moles} = \frac{\text{Mass of Substance (g)}}{\text{Molar Mass (g/mol)}}ewlineewlineewlineewlineewlineewlineewlineewline\end{equation}\]This step is the starting point for most stoichiometric calculations and allows you to transition from mass to mole, a more useful unit for reaction calculations.

Net Ionic Equations

In chemistry, reactions often occur in solutions where ions are the main players. While full molecular equations can be written, they often include spectator ions that do not participate in the reaction. Net ionic equations simplify this by only showing the species that actually change during the reaction.

Creating a net ionic equation involves the following steps:

• Writing the balanced molecular equation.
• Dissociating any strong electrolytes into ions.
• Identifying and removing spectator ions that appear unchanged on both sides of the equation.
• Writing the remaining species to represent the actual chemical change.

In our exercise, the net ionic equation for the precipitation of silver chloride from aqueous solutions of silver nitrate and sodium chloride is simply:\[\begin{equation}\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq) \rightarrow \mathrm{AgCl}(s)ewlineewlineewlineewlineewlineewlineewlineewline\end{equation}\]This depicts the formation of a solid precipitate from the soluble ions, removing spectator ions like Na⁺ and NO₃⁻.

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds in solution form an insoluble solid, known as the precipitate. This type of reaction is easily recognizable by the formation of a solid from a previously clear solution.

The ability to predict whether a reaction will result in a precipitate depends on the solubility rules. Some combinations of ions yield insoluble compounds. For example, silver chloride (AgCl) is typically insoluble in water, making it likely to form a precipitate when silver (Ag⁺) and chloride (Cl⁻) ions meet in solution.

To verify the completion of a precipitation reaction, stoichiometry can be used to compare the moles of reactants. The ratio in which reactants combine, a concept pivotal to stoichiometry, determines whether one reactant will be in excess or completely consumed, as seen in the exercise when comparing amounts of sodium chloride and silver nitrate.