Question

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

Short answer

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

Step-by-step solution

Write the balanced chemical equation

We have the precipitation reaction between sodium chromate (\(\mathrm{Na}_{2} \mathrm{CrO}_{4}\)) and silver nitrate (\(\mathrm{AgNO}_{3}\)). In this reaction, the silver ions will form a precipitate with chromate ions, resulting in the formation of silver chromate (\(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)). The balanced chemical equation for this reaction is: $$2\,\mathrm{AgNO}_{3} \,+\, \mathrm{Na}_{2}\mathrm{CrO}_{4} \,\longrightarrow\, \mathrm{Ag}_{2}\mathrm{CrO}_{4} \,+\, 2\,\mathrm{NaNO}_{3}$$

Calculate the moles of silver ions in the solution

Next, we will determine the moles of silver ions in the \(\mathrm{AgNO}_{3}\) solution. To do this, we use the volume and concentration of the solution. Moles = Volume × Concentration The volume of the solution is 75.0 mL, which we must convert to liters. 1 L = 1000 mL, so we have: $$\text{Volume} = \frac{75.0\,\text{mL}}{1000\,\text{mL/L}} = 0.0750 \,\text{L}$$ Now, we can calculate the moles of \(\mathrm{AgNO}_{3}\): Moles of \(\mathrm{AgNO}_{3}\) = 0.0750 L × 0.100 mol/L = 0.00750 mol

Determine the moles of sodium chromate needed

Using the stoichiometry from the balanced chemical equation, we can determine the moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) needed to react with the moles of \(\mathrm{AgNO}_{3}\). From the balanced equation, we can see that one mole of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to react with two moles of \(\mathrm{AgNO}_{3}\). Therefore, we can set up the following proportion to find the moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required: $$\frac{\text{moles of }\mathrm{Na}_{2}\mathrm{CrO}_{4}}{0.00750\,\text{mol } \mathrm{AgNO}_{3}} = \frac{1\,\text{mol }\mathrm{Na}_{2}\mathrm{CrO}_{4}}{2\,\text{mol}\,\mathrm{AgNO}_{3}}$$ We can solve for the moles of \(\mathrm{Na}_{2}\mathrm{CrO}_{4}\): $$\text{moles of }\mathrm{Na}_{2}\mathrm{CrO}_{4} = 0.00750\,\text{mol }\mathrm{AgNO}_{3} × \frac{1\,\text{mol}\,\mathrm{Na}_{2}\mathrm{CrO}_{4}}{2\,\text{mol }\mathrm{AgNO}_{3}} = 0.00375\,\text{mol }\mathrm{Na}_{2}\mathrm{CrO}_{4}$$

Calculate the mass of sodium chromate required

Finally, we can determine the mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required to react with the moles of silver ions present. We will use the molar mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) to convert moles to grams. $$\text{Mass of }\mathrm{Na}_{2}\mathrm{CrO}_{4} =\text{Moles }× \text{Molar mass}$$ The molar mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is: $$2\,\text{mol Na }(22.99\,\text{g/mol}) + 1\,\text{mol Cr }(51.99\,\text{g/mol}) + 4\,\text{mol O }(16.00\,\text{g/mol}) = 161.97\,\text{g/mol}$$ Now we can find the mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required: $$\text{Mass of }\mathrm{Na}_{2}\mathrm{CrO}_{4} = 0.00375\,\text{mol} × 161.97\,\text{g/mol} = 0.607\,\text{g}$$ So, a mass of 0.607 g of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from 75.0 mL of 0.100-M solution of \(\mathrm{AgNO}_{3}\).

Key concepts

Chemical Reactions

Chemical reactions involve substances, known as reactants, interacting to form new substances, called products. In this context, a precipitation reaction is a type of chemical reaction where two soluble salts react in solution to form an insoluble solid called a precipitate.

For example, when sodium chromate (\(\mathrm{Na}_{2} \mathrm{CrO}_{4}\)) is mixed with silver nitrate (\(\mathrm{AgNO}_{3}\)), they undergo a double displacement reaction. This reaction will swap components between the reactants, resulting in the formation of silver chromate (\(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)) and sodium nitrate (\(\mathrm{NaNO}_{3}\)).

The balanced chemical equation for this reaction is:

• \(2\,\mathrm{AgNO}_{3} \,\,+\,\, \mathrm{Na}_{2}\mathrm{CrO}_{4} \,\longrightarrow\, \mathrm{Ag}_{2}\mathrm{CrO}_{4} \,\,+\,\, 2\,\mathrm{NaNO}_{3}\)

The equation tells us the proportion in which substances react or are produced. Here, 2 moles of silver nitrate react with 1 mole of sodium chromate to form 1 mole of silver chromate and 2 moles of sodium nitrate.

Understanding the balanced equation is crucial. It reveals the stoichiometric relationships between different reactants and products, which will guide precise calculations later on.

Mole Calculations

Mole calculations are fundamental in stoichiometry, providing a way to convert between the mass of a substance and the number of particles it contains. A mole is a unit representing \(6.022 \times 10^{23}\) particles (atoms, molecules).

When we calculate moles, we convert given volumes or masses into moles using either molarity or molar mass. In the example exercise, we want to determine how much sodium chromate is needed to completely precipitate silver ions from a solution.

To find out, start by figuring out the moles of silver nitrate available. Use the molarity of the solution along with its volume:

• \(0.100\,\mathrm{M} \times 0.0750\,\mathrm{L} = 0.00750\,\text{mol } \mathrm{AgNO}_{3}\).

The balanced equation shows that 1 mole of sodium chromate reacts with 2 moles of silver nitrate. Therefore, divide the moles of silver nitrate by 2 to find the moles of sodium chromate needed:

• \(0.00750\,\text{mol}\,\mathrm{AgNO}_{3}\times \frac{1\,\text{mol}\,\mathrm{Na}_{2}\mathrm{CrO}_{4}}{2\,\text{mol }\,\mathrm{AgNO}_{3}} = 0.00375\,\text{mol }\,\mathrm{Na}_{2}\mathrm{CrO}_{4}\).

This mole calculation is key for advancing to the next step — converting moles to mass using molar mass.

Precipitation Reactions

Precipitation reactions are exciting processes where substances in solution react to create a solid, known as a precipitate. This can often be seen as a cloudy appearance or solid chunks forming within the solution.

In the case study, silver chromate (\(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)) forms as the precipitate. This occurs when silver ions from silver nitrate combine with chromate ions from sodium chromate:

• \(2\mathrm{Ag}^{+}\, \text{(from: }\,\mathrm{AgNO}_{3}\text{)} + \mathrm{CrO}_{4}^{2-}\, \text{(from: }\,\mathrm{Na}_{2}\mathrm{CrO}_{4}\text{)} \rightarrow \mathrm{Ag}_{2}\mathrm{CrO}_{4} \text{ (precipitate)}\)

Understanding precipitation reactions requires recognizing that these reactions require proper conditions where the products are insoluble in water.

Examining solubility rules or utilizing a solubility chart can forecast whether a precipitate will form upon mixing two aqueous solutions.

These reactions are crucial.... Especially in fields like analytical chemistry, as they help separate and identify components within a mixture.