Question

Barium chloride solutions are used in chemical analysis for the quantitative precipitation of sulfate ion from solution. $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{BaSO}_{4}(s) $$ Suppose a solution is known to contain on the order of \(150 \mathrm{mg}\) of sulfate ion. What mass of barium chloride should be added to guarantee precipitation of all the sulfate ion?

Short answer

The mass of barium chloride (\(\mathrm{BaCl}_{2}\)) required to precipitate all the sulfate ion (\(\mathrm{SO}_{4}^{2-}\)) present in the solution (150 mg) is 0.325 g.

Step-by-step solution

Write down the balanced chemical equation

In this case, the balanced chemical equation is already provided: \[ \mathrm{Ba}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq) \rightarrow \mathrm{BaSO}_{4}(s) \]

Determine the number of moles of sulfate ions

Given that there are 150 mg of sulfate ions, we need to convert this mass into moles. First, find the molar mass of sulfate ion (SO₄²⁻): Molar mass of S = 32.06 g/mol Molar mass of O = 16.00 g/mol Molar mass of SO₄²⁻ = 32.06 + (4 × 16.00) = 96.06 g/mol Now, convert the mass of sulfate (150 mg) to moles: Mass of sulfate = 150 mg = 0.150 g Number of moles of sulfate ions = mass / molar mass Number of moles of sulfate ions = \( \frac{0.150 \,\text{g}}{96.06\, \text{g/mol}} \) = 1.56 × 10⁻³ moles

Use the stoichiometry of the balanced equation to determine the moles of barium chloride needed

From the balanced equation, the mole ratio of \(\mathrm{Ba}^{2+}\) ion to \(\mathrm{SO}_{4}^{2-}\) ion is 1:1. Therefore, the moles of \(\mathrm{Ba}^{2+}\) ions required are equal to the moles of sulfate ions: Moles of \(\mathrm{Ba}^{2+}\) ions = 1.56 × 10⁻³ moles Since barium chloride (\(\mathrm{BaCl}_{2}\)) provides the \(\mathrm{Ba}^{2+}\) ions, the moles of \(\mathrm{BaCl}_{2}\) required will be equal to the moles of \(\mathrm{Ba}^{2+}\) ions needed. Moles of \(\mathrm{BaCl}_{2}\) = 1.56 × 10⁻³ moles

Convert moles of barium chloride to mass

Now, we need to find the mass of the barium chloride required. First, determine the molar mass of barium chloride (\(\mathrm{BaCl}_{2}\)): Molar mass of Ba = 137.33 g/mol Molar mass of Cl = 35.45 g/mol Molar mass of \(\mathrm{BaCl}_{2}\) = 137.33 + (2 × 35.45) = 208.23 g/mol Next, use the number of moles of \(\mathrm{BaCl}_{2}\) to find its mass: Mass of \(\mathrm{BaCl}_{2}\) = moles × molar mass Mass of \(\mathrm{BaCl}_{2}\) = (1.56 × 10⁻³ moles) × (208.23 g/mol) = 0.325 g Thus, the mass of barium chloride required to precipitate all the sulfate ion present in the solution (150 mg) is 0.325 g.

Key concepts

Chemical Reaction

A chemical reaction involves the transformation of one or more substances into new products. This process is depicted using chemical equations, which provide a symbolic representation of the reactants and the products. For example, in the reaction \( \mathrm{Ba}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq) \rightarrow \mathrm{BaSO}_{4}(s) \), barium ions and sulfate ions react to form solid barium sulfate. This is a prime example of a precipitation reaction, where the formation of a solid from aqueous ions occurs.

• Reactants: Substances that undergo change, in this case, \( \mathrm{Ba}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \).
• Products: New substances formed, in this scenario \( \mathrm{BaSO}_{4} \).

Chemical reactions are governed by the principle of conservation of mass, which means the total mass of the products equals that of the reactants. This concept is crucial in quantitative analysis, as it ensures accurate measurement and calculation of the components involved.

Moles Calculation

Calculating moles is a pivotal part of stoichiometry, which deals with quantitative relationships in chemical equations. To perform moles calculation, you need the mass of the substance and its molar mass (the mass of one mole of a substance).First, determine the molar mass. For sulfate \( \text{SO}_4^{2-} \), the molar mass is calculated based on its atomic components.

• Sulfur (S): 32.06 g/mol
• Oxygen (O): 16.00 g/mol
• Molar mass of \( \text{SO}_4^{2-} \): 96.06 g/mol

Next, convert the provided mass to moles:\[\text{Moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}}\]Given a sulfate mass of 0.150 grams, the number of moles is:\[\frac{0.150 \text{g}}{96.06 \text{g/mol}} = 1.56 \times 10^{-3} \text{ moles}\]This straightforward calculation is essential in determining the quantities required in a chemical reaction.

Precipitation Reaction

A precipitation reaction involves the formation of an insoluble solid from the reaction of two soluble substances in a solution. In our example, the reaction between \( \text{Ba}^{2+} \) ions and \( \text{SO}_4^{2-} \) ions produces barium sulfate (\( \text{BaSO}_4 \)), a solid precipitate.

• Occurs when product is insoluble in water.
• Common in qualitative analysis, helping determine presence of specific ions.

In our reaction:\( \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s) \)The reaction stops when either of the reacting ions is completely used, which helps in controlling the purity and yield of the desired product. This reaction is a practical way of removing ions from solutions, showing the utility of stoichiometry in industrial and laboratory processes.

Quantitative Analysis

Quantitative analysis refers to the measurement of the amount or concentration of substances in a sample. In the context of our chemical reaction, it involves calculating the precise amount of barium chloride needed to completely precipitate sulfate ions from a solution.Start by using stoichiometry to understand the mole relationships between reactants and products. The balanced equation shows a 1:1 molar ratio:

• \( \text{Moles of } \text{Ba}^{2+} = \text{Moles of } \text{SO}_4^{2-} \)

Apply this ratio to determine the exact amount of barium chloride required:

• Given 1.56 \times 10^{-3} moles of sulfate, equivalent moles of \( \text{BaCl}_2 \) are needed.

Finally, convert moles to mass using the molar mass of \( \text{BaCl}_2 \) (208.23 g/mol):\[\text{Mass} = 1.56 \times 10^{-3} \text{ moles} \times 208.23 \text{ g/mol} = 0.325 \text{ g}\]This calculated mass of barium chloride ensures complete reaction with the sulfate ions, showcasing the precision of quantitative analysis.