Question

How many grams of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) are required to precipitate all the sulfate ion present in \(15.3 \mathrm{mL}\) of 0.139 \(M \mathrm{H}_{2} \mathrm{SO}_{4}\) solution? \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q)\)

Short answer

To find the grams of \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2}\) required to precipitate all the sulfate ions present in the given solution, first calculate the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) in the solution as follows: \( \mathrm{moles} \ \mathrm{of} \ \mathrm{H}_{2}\mathrm{SO}_{4} = (0.139 \ \mathrm{M})(0.0153 \ \mathrm{L}) = 0.00213 \ \mathrm{mol} \) Since the stoichiometry between \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2}\) and \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is 1:1, the moles of \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2}\) needed is also 0.00213 mol. Calculate the molar mass of \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2}\): Molar mass of \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2} = 137.33 + 2\times(14.01+3\times16) = 261.34 \ \mathrm{g \ mol^{-1}}\) Finally, find the grams of \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2}\) required: \( \mathrm{grams \ of \ Ba(NO}_{3}\mathrm{)}_{2} = (0.00213 \ \mathrm{mol}) \times (261.34 \ \mathrm{g \ mol^{-1}}) = 0.557 \ \mathrm{g} \) Thus, 0.557 grams of \(\mathrm{Ba(NO}_{3}\mathrm{)}_{2}\) are required to precipitate all the sulfate ions present in the given solution.

Step-by-step solution

Find the moles of sulfate ions from the H₂SO₄ solution

Given the concentration of the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution (0.139 M) and the volume (15.3 mL), we can find the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and therefore the moles of sulfate ions. First, we need to convert the volume from milliliters to liters: \( 15.3 \ \mathrm{mL} \times \frac{1 \ \mathrm{L}}{1000 \ \mathrm{mL}} = 0.0153 \ \mathrm{L} \) Now, we can find the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) \( \mathrm{moles} \ \mathrm{of} \ \mathrm{H}_{2}\mathrm{SO}_{4} = \mathrm{concentration} \times \mathrm{volume} \) \( \mathrm{moles} \ \mathrm{of} \ \mathrm{H}_{2}\mathrm{SO}_{4} = (0.139 \ \mathrm{M})(0.0153 \ \mathrm{L}) \)

Calculate the moles of Ba(NO₃)₂ needed

Using the balanced chemical equation: \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q)\) We observe that the stoichiometry between \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is 1:1. Hence, the moles of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) needed is equal to the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\).

Convert the moles of Ba(NO₃)₂ to grams

The molar mass of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) can be found as follows: Molar mass of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} = 137.33 + 2\times(14.01+3\times16)\) (Using the molar masses of Ba, N, and O) Now, we can find the mass of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) required by multiplying the moles with its molar mass: \( \mathrm{grams \ of \ Ba(NO}_{3}\mathrm{)}_{2} = \mathrm{moles \ of} \ \mathrm{Ba(NO}_{3}\mathrm{)}_{2} \times \mathrm{molar \ mass \ of} \ \mathrm{Ba(NO}_{3}\mathrm{)}_{2} \) Combine all these steps to find the answer.

Key concepts

Molecular Weight

Molecular weight, also known as molar mass, is essential for converting moles of a substance into grams and vice versa. To find the molecular weight of a compound, you sum the atomic masses of each element in the compound, multiplied by the number of atoms of that element present. For example, the molecular weight of barium nitrate, \( \mathrm{Ba(NO}_3 \mathrm{)}_2 \), can be calculated using the atomic masses from the periodic table:

• Barium (Ba) has an atomic mass of 137.33 g/mol.
• Nitrogen (N) has an atomic mass of 14.01 g/mol.
• Oxygen (O) has an atomic mass of 16.00 g/mol.

Barium nitrate contains one barium atom, two nitrogen atoms, and six oxygen atoms. Thus, the molar mass computation would look like this: \( 137.33 + 2 \times (14.01 + 3\times16) \). This exercise illustrates how molecular weight connects the mole concept to practical mass measurements used in the lab.

Molarity

Molarity (M) is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The given exercise provides the molarity and volume of sulfuric acid (\( \mathrm{H}_2 \mathrm{SO}_4 \)) solution, which is essential for determining how many moles of \( \mathrm{SO}_4^{2-} \) ions are present. To find the number of moles, you use the formula:
\[ \text{Moles of } \mathrm{H}_2 \mathrm{SO}_4 = \text{Molarity} \times \text{Volume (in L)} \].
Considering the viscous nature of chemistry, remember to convert mL to L by dividing by 1000. Ultimately, this helps in stoichiometry calculations to figure out how much of another reactant is needed in a chemical reaction.

Balancing Chemical Equations

Balancing chemical equations is vital because it ensures the law of conservation of mass is satisfied; atoms on the reactant side must equal those on the product side. In the exercise's reaction:
\( \mathrm{Ba(NO}_3\mathrm{)}_2 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + 2\mathrm{HNO}_3 \),
the equation is already balanced.
Each type of atom found on the left has an equivalent and equal count on the right. This balanced equation tells us that one mole of \( \mathrm{Ba(NO}_3\mathrm{)}_2 \) reacts with one mole of \( \mathrm{H}_2 \mathrm{SO}_4 \). The stoichiometric relationships can be used to calculate reactant or product amounts, as seen in the solution, where moles of \( \mathrm{Ba(NO}_3\mathrm{)}_2 \) correspond directly with moles of \( \mathrm{SO}_4^{2-} \) ions present.

Precipitation Reaction

A precipitation reaction occurs when two solutions are mixed and an insoluble solid, known as a precipitate, forms. In our exercise, when \( \mathrm{Ba(NO}_3 \mathrm{)}_2 \) reacts with \( \mathrm{H}_2 \mathrm{SO}_4 \), barium sulfate \( \mathrm{BaSO}_4 \) is the precipitate formed. Precipitation reactions often involve ionic compounds, where cations and anions rearrange to create an insoluble product in water. This reaction's core concept helps us predict the formation of solids from solutions, vital for applications in laboratory and industrial chemistry. Recognizing precipitation reactions allows identification of products that will settle out of solution, affecting stability and solubility.