Question

What volume (in \(\mathrm{mL}\) ) of \(0.25 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution is needed to precipitate all the barium, as \(\mathrm{BaSO}_{4}(s)\) from \(12.5 \mathrm{mL}\) of \(0.15 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution? \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow_{\mathrm{BaSO}_{4}(s)+2 \mathrm{NaNO}_{3}(a q)}\)

Short answer

The required volume of \(0.25 M\) Na2SO4 solution to precipitate all the barium as BaSO4 from \(12.5 mL\) of \(0.15 M\) Ba(NO3)2 solution is \(7.5 mL\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is already given in the problem: Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaNO3(aq) Make sure to understand the stoichiometry in this reaction: 1 mole of Ba(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4 and 2 moles of NaNO3.

Calculate moles of Ba(NO3)2

We have the volume and concentration of the Ba(NO3)2 solution, which is 12.5 mL and 0.15 M, respectively. To find the moles of Ba(NO3)2, multiply the volume by the concentration: moles of Ba(NO3)2 = volume (L) × concentration (M) moles of Ba(NO3)2 = \(12.5 \times 10^{-3} L \times 0.15 M\) moles of Ba(NO3)2 = \(1.875 \times 10^{-3} mol\)

Calculate moles of Na2SO4 needed

According to the balanced chemical equation, 1 mole of Ba(NO3)2 reacts with 1 mole of Na2SO4. So, we need the same moles of Na2SO4 as that of Ba(NO3)2. moles of Na2SO4 needed = moles of Ba(NO3)2 moles of Na2SO4 needed = \(1.875 \times 10^{-3} mol\)

Calculate the volume of Na2SO4 solution needed

We have the concentration of the Na2SO4 solution (0.25 M) and the moles of Na2SO4 required. To find the volume, divide moles of Na2SO4 by its concentration: volume of Na2SO4 solution (L) = \( \frac{moles \: of \: Na_{2}SO_{4}}{concentration \: of \: Na_{2}SO_{4} \: solution} \) volume of Na2SO4 solution (L) = \( \frac{1.875 \times 10^{-3} mol}{0.25 M} \) volume of Na2SO4 solution (L) = \(7.5 \times 10^{-3} L\) Now switch the volume of Na2SO4 solution to milliliters: volume of Na2SO4 solution (mL) = \(7.5 \times 10^{-3} L \times \frac{1000 mL}{1 L}\) volume of Na2SO4 solution (mL) = 7.5 mL

Final Answer

The required volume of 0.25 M Na2SO4 solution to precipitate all the barium as BaSO4 from 12.5 mL of 0.15 M Ba(NO3)2 solution is 7.5 mL.

Key concepts

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid. This solid is called a precipitate. In our given problem, the precipitation reaction happens between sodium sulfate (\(\text{Na}_2\text{SO}_4\)) and barium nitrate (\(\text{Ba(NO}_3)_2\)). When these solutions are mixed together, they react to form barium sulfate (\(\text{BaSO}_4\)), the precipitate, and sodium nitrate (\(\text{NaNO}_3\)), which remains in solution.- **Key aspects of a precipitation reaction:** - **Formation of a solid (precipitate):** Barium sulfate (\(\text{BaSO}_4\)) in this case. - **Reactants' solubility:** Most reactants are soluble before they react. - **Insolubility of the product:** The precipitate does not dissolve in the solution, hence it settles out.Understanding this reaction helps to predict the result of mixing substances and confirm the presence of certain ions based on whether a visible solid forms.

Chemical Equation Balancing

Balancing chemical equations is a crucial skill in chemistry that ensures the conservation of mass. In a balanced chemical equation, the number of atoms for each element is the same on both the reactant and product sides. This reflects the law of conservation of matter that matter is neither created nor destroyed in a chemical reaction. For the chemical equation:\[\text{Ba(NO}_3)_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaNO}_3(aq)\]- **Steps to balance the equation:** - **Count each type of atom on both sides** of the arrow. - **Adjust coefficients** (the numbers in front of chemical formulas) to make the numbers equal.In this exercise, the equation is already balanced:- 1 mole of barium nitrate reacts with 1 mole of sodium sulfate.- This produces 1 mole of barium sulfate and 2 moles of sodium nitrate.Balancing ensures the stoichiometry needed for calculating reactants and products during experiments and calculations.

Molar Concentration

Molar concentration, or molarity, is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution (mol/L). In your exercise, you have to work with sodium sulfate and barium nitrate's different molar concentrations.- **To calculate molarity:** - **Formula:** \[ M = \frac{n}{V} \] where \(M\) is molarity, \(n\) is the number of moles, and \(V\) is the volume in liters.In the problem:- Sodium sulfate and barium nitrate solutions have different molarities (\(0.25 M\) and \(0.15 M\) respectively).- Calculating with molarity involves adjusting volumes and concentrations, which is key in reactions as it dictates how much is needed for reactions to go to completion.This concept allows accurate preparation of solutions for various reactions by ensuring the right amounts of reactants are used.

Solution Volume Calculation

Solution volume calculation is necessary to determine the amount of a solution needed in a reaction. It often involves using molarity and the number of moles of a solute.In our exercise, we want to determine how much \(\text{Na}_2\text{SO}_4\) solution is needed to fully react with a given volume of \(\text{Ba(NO}_3)_2\) solution. Here's how you do it:- **Given data:** - Moles of \(\text{Ba(NO}_3)_2\): \(1.875 \times 10^{-3} \text{ mol}\). - Molarity of \(\text{Na}_2\text{SO}_4\): \(0.25 M\).- **Volume calculation formula:** \[ \text{Volume} = \frac{\text{moles of } \text{Na}_2\text{SO}_4}{\text{concentration}} = \frac{1.875 \times 10^{-3}}{0.25} = \text{in liters} \]- **Convert to milliliters** (since practical lab measurements): \(\text{Volume in mL} = 7.5 \text{ mL}\).This calculation ensures you use the exact necessary volume to react completely, minimizing waste or excess.