Problem 146 - General Chemistry

Question

A \(25.0-\mathrm{mL}\) sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution to produce barium sulfate crystals, which were filtered from the solution. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{BaSO}_{4}(s)\) If \(5.719 \mathrm{~g}\) of barium sulfate was obtained, what was the molarity of the original \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Step-by-Step Solution

Answer

The molarity of the original sodium sulfate solution is 0.980 M.

Step 1: Calculate Moles of Barium Sulfate

First, determine the number of moles of barium sulfate (\(\mathrm{BaSO}_{4}\)) formed. Use the formula for calculating moles from mass:\[moles = \frac{\text{mass}}{\text{molar mass}}\]The molar mass of \(\mathrm{BaSO}_{4}\) is calculated as:\[137.33\,(\text{Ba}) + 32.07\,(\text{S}) + 4 \times 16.00\,(\text{O}) = 233.39\,\text{g/mol}\]Therefore, the moles of \(\mathrm{BaSO}_{4}\) is:\[moles = \frac{5.719\,\text{g}}{233.39\,\text{g/mol}} = 0.0245\,\text{mol}.\]

Step 2: Determine Moles of Sodium Sulfate

The chemical equation shows that one mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) reacts with one mole of \(\mathrm{BaCl}_{2}\) to produce one mole of \(\mathrm{BaSO}_{4}\). Therefore, the number of moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is equal to the number of moles of \(\mathrm{BaSO}_{4}\):\[0.0245\,\text{mol}\,\mathrm{Na}_{2} \mathrm{SO}_{4} = 0.0245\,\text{mol}\,\mathrm{BaSO}_{4}.\]

Step 3: Calculate Molarity of Sodium Sulfate Solution

Molarity (\(M\)) is defined as the moles of solute divided by the volume of the solution in liters. Convert the volume of the sodium sulfate solution from milliliters to liters:\[25.0\,\text{mL} = 0.0250\,\text{L}\]Now, calculate the molarity:\[M = \frac{0.0245\,\text{mol}}{0.0250\,\text{L}} = 0.980\,\text{M}\]

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It tells you the proportions of each ingredient needed for the reaction to take place. Imagine baking a cake without knowing how much flour, sugar, or eggs to use. Similar confusion can occur in chemistry without stoichiometry.
In the chemical reaction given, \[\mathrm{Na}_2 \mathrm{SO}_4(aq) + \mathrm{BaCl}_2(aq) \rightarrow 2\mathrm{NaCl}(aq) + \mathrm{BaSO}_4(s)\] stoichiometry helps us determine how much of a substance is needed or produced. The coefficients (the numbers before chemical formulas) tell us the ratio of molecules that react and are formed.

• For every 1 mole of \(\mathrm{Na}_2 \mathrm{SO}_4\), 1 mole of \(\mathrm{BaSO}_4\) is produced.
• This 1:1 ratio simplifies the calculation of reactants and products.

So by using stoichiometry, we can accurately predict how much \(\mathrm{BaSO}_4\) will be produced from a given amount of \(\mathrm{Na}_2 \mathrm{SO}_4\). It's like knowing the exact ingredients needed to get a perfect dish every time.

Moles Calculation

Moles calculation is crucial in chemistry because it links the microscopic world of atoms and molecules to the macroscopic world we can observe and measure. A mole is simply a unit that represents \(6.022 \times 10^{23}\) particles (atoms, molecules, etc.). This number is called Avogadro's number. To calculate moles, we use the formula: \[moles = \frac{\text{mass}}{\text{molar mass}}\]Let's break down the calculation:

• The molar mass is the mass of one mole of a substance, measured in grams per mole \(\text{g/mol}\).
• By dividing the mass you have by the molar mass, you get the number of moles.

For example, in the given exercise, we calculated the moles of \(\mathrm{BaSO}_4\) as follows:\[moles = \frac{5.719\,\text{g}}{233.39\,\text{g/mol}} = 0.0245\,\text{mol}.\]This shows the conversion from grams (something we can measure) to moles (something we use to predict outcomes in a reaction). It’s like converting currency; you may know how many dollars you have, but converting them to euros may be necessary for completing a transaction elsewhere.

Chemical Reaction Equations

Chemical reaction equations paint a clear picture of how substances interact. They show the starting materials (reactants) and the resulting products in a chemical reaction. These equations need to be balanced, meaning the numbers of each type of atom on both sides must be equal. This follows the Law of Conservation of Mass, stating that mass is neither created nor destroyed in a chemical reaction. In the given equation:\[\mathrm{Na}_2 \mathrm{SO}_4(aq) + \mathrm{BaCl}_2(aq) \rightarrow 2\mathrm{NaCl}(aq) + \mathrm{BaSO}_4(s)\]

• Reactants: \(\mathrm{Na}_2 \mathrm{SO}_4\) and \(\mathrm{BaCl}_2\)
• Products: \(\mathrm{NaCl}\) and \(\mathrm{BaSO}_4\)

Understanding these components provides the basis for predicting what will happen in a reaction and how much of each substance is involved. It’s like following directions: turn left at the stop sign, and you'll arrive at your destination. Each symbol and number in the equation directs the outcome of the chemical reaction, ensuring things go as planned.