Question

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{~mL}\) of a \(0.100 M\) solution of \(\mathrm{AgNO}_{3}\) ?

Short answer

To precipitate all the silver ions from a \(75.0\,\mathrm{mL}\) solution of \(0.100\, \mathrm{M}\, \mathrm{AgNO}_{3}\), approximately \(0.607\, \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required.

Step-by-step solution

Write the balanced chemical equation

The reaction that occurs when sodium chromate (Na₂CrO₄) reacts with silver nitrate (AgNO₃) to precipitate silver ions (Ag⁺) can be described by the following balanced chemical equation: \[2 AgNO_3(aq) + Na_2CrO_4(aq) \rightarrow Ag_2CrO_4(s) + 2 NaNO_3(aq)\]

Find the moles of AgNO₃ in the given solution

We are given the volume (\(75.0\,\text{mL}\)) and concentration (\(0.100\,\text{M}\)) of the AgNO₃ solution. We can use these values to calculate the moles of AgNO₃ in the solution. Moles = Molarity × Volume The volume needs to be converted to liters, so \(\text{Volume} = 75.0\,\text{mL} \times \frac{1 \,\text{L}}{1000\, \text{mL}} = 0.075\, \text{L}\). Then, Moles of AgNO₃ = \(0.100\, \text{M} \times 0.075\, \text{L} = 0.0075\, \text{mol}\)

Find the moles of Na₂CrO₄ required to precipitate all the Ag⁺ ions

From the balanced chemical equation, we can see that 2 moles of AgNO₃ react with 1 mole of Na₂CrO₄ to form 2 moles of NaNO₃ and 1 mole of Ag₂CrO₄. So, the mole ratio of AgNO₃ to Na₂CrO₄ is 2:1. To find the moles of Na₂CrO₄ required to precipitate all the silver ions, we can use this mole ratio: Moles of Na₂CrO₄ = \(\text{Moles of AgNO₃} \, \times \, \frac{1\, \text{mol Na₂CrO₄}}{2\, \text{mol AgNO₃}} = 0.0075\, \text{mol} \, \times \, \frac{1}{2} = 0.00375\, \text{mol}\)

Calculate the mass of Na₂CrO₄ needed

Now that we know the moles of Na₂CrO₄ required, we can calculate the mass of Na₂CrO₄ using its molar mass. The molar mass of Na₂CrO₄ is approximately \(2(22.99) + 51.996 + 4(16.00) = 161.984\, \text{g/mol}\). Then, Mass of Na₂CrO₄ = \(\text{Moles of Na₂CrO₄} \, \times \, \text{Molar mass of Na₂CrO₄} = 0.00375\, \text{mol} \, \times \, 161.984\, \text{g/mol} = 0.607\, \text{g}\) So, the mass of Na₂CrO₄ required to precipitate all the silver ions from a \(75.0\,\mathrm{mL}\) solution of \(0.100\, \mathrm{M}\, \mathrm{AgNO}_{3}\) is approximately \(0.607\, \mathrm{g}\).

Key concepts

Moles Calculation

In chemistry, moles serve as a bridge between the atomic world and the macroscopic amounts we handle in the lab. To perform moles calculation, it is important to understand that the "mole" relates to Avogadro's number, which is approximately \(6.022 \times 10^{23}\). This number signifies the quantity of atoms, ions, or molecules in one mole of a substance.
To find moles in a solution, you use the relation:

• Moles = Molarity \(\times\) Volume

The molarity gives the concentration of a solution, expressed as moles per liter, and the volume must be in liters to ensure consistency. For example, if you have 0.100 M (molar) solution and 75.0 mL of this solution, you first convert the volume to liters by dividing by 1000, giving 0.075 L. Then, multiplying the molarity by this volume provides the number of moles of the solution, which in our case of AgNO₃, is \(0.100\,\text{M} \times 0.075\,\text{L} = 0.0075\,\text{mol}\).
Understanding moles calculation is crucial as it lets you quantify chemicals in practical laboratory scenarios, facilitating further calculations like mass or concentration conversions.

Balanced Chemical Equation

The balanced chemical equation is essential in any chemical reaction analysis as it provides the stoichiometric relationship between the reactants and products. This balance respects the Law of Conservation of Mass, meaning all atoms present in the reactants must be accounted for in the products.
When sodium chromate \((\text{Na}_2\text{CrO}_4)\) reacts with silver nitrate \((\text{AgNO}_3)\), the products formed are silver chromate \((\text{Ag}_2\text{CrO}_4)\) and sodium nitrate \((\text{NaNO}_3)\). Our balanced equation:

• \[2 \text{AgNO}_3 (aq) + \text{Na}_2\text{CrO}_4 (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s) + 2 \text{NaNO}_3 (aq)\]

highlights the stoichiometry required: two moles of silver nitrate react with one mole of sodium chromate. Balancing equations requires adjusting the coefficients before each compound until the same number of each type of atom appears on both sides of the equation. This equation ensures accurate calculations for further investigations and predicts the amounts of reactants needed or products formed in a reaction, enhancing the efficiency of experimental setups.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction occurring when two soluble ions in separate solutions combine to form an insoluble compound, which settles from the solution as a solid, known as a precipitate.
In our case, when sodium chromate \((\text{Na}_2\text{CrO}_4)\) and silver nitrate \((\text{AgNO}_3)\) are mixed, the silver ions \((\text{Ag}^+)\) react with chromate ions \((\text{CrO}_4^{2-})\) to form silver chromate \((\text{Ag}_2\text{CrO}_4)\), a bright red precipitate.

• This process is key in removing or separating components from a solution.
• The reaction: \[2 \text{AgNO}_3 (aq) + \text{Na}_2\text{CrO}_4 (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s) + 2 \text{NaNO}_3 (aq)\]

shows the formation of a precipitate of \(\text{Ag}_2\text{CrO}_4\). Precipitation reactions are crucial in various applications such as water purification, chemical analysis, and manufacturing processes where specific substances need to be extracted from mixtures.