Question

In a laboratory, \(6.82 \mathrm{~g}\) of \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) is dissolved in enough water to form \(0.500 \mathrm{~L}\) of solution. A \(0.100\) -L sample is withdrawn from this stock solution and titrated with a \(0.0245 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\). What volume of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) solution is needed to precipitate all the \(\mathrm{Sr}^{2+}(a q)\) as \(\mathrm{SrCrO}_{4} ?\)

Short answer

To precipitate all the Sr\(^{2+}\)(aq) ions as SrCrO\(_4\), you will need \(0.2629 \mathrm{~L}\) or \(262.9 \mathrm{~mL}\) of the \(0.0245 \mathrm{~M}\) Na\(_2\)CrO\(_4\) solution.

Step-by-step solution

Find the moles of Sr\(^{2+}\) ions

First, we need to calculate the moles of Sr\(^{2+}\) ions in the 0.100 L withdrawn sample. Using the mass and volume of the Sr(NO\(_3\))\(_2\) stock solution, we can calculate its concentration: Moles of Sr(NO\(_3\))\(_2\) = mass / molar mass Sr: 87.62 g/mol, N: 14.01 g/mol, O: 16.00 g/mol Molar mass of Sr(NO\(_3\))\(_2\) = 1 * 87.62 + 2 * (1 * 14.01 + 3 * 16.00) = 211.63 g/mol Moles of Sr(NO\(_3\))\(_2\) = 6.82 g / 211.63 g/mol = 0.0322 mol Concentration of Sr(NO\(_3\))\(_2\) = moles/volume = 0.0322 mol / 0.5 L = 0.0644 M Next, using the concentration, we will calculate the moles of Sr\(^{2+}\) ions in the 0.100 L withdrawn sample: Moles of Sr\(^{2+}\) = concentration × volume = 0.0644 M × 0.1 L = 0.00644 mol

Calculate the moles of Na\(_2\)CrO\(_4\) required

We'll now use the stoichiometry of the reaction to determine the moles of Na\(_2\)CrO\(_4\) required to precipitate all the Sr\(^{2+}\) ions: Sr\(^{2+}\) + CrO\(_4^{2-}\) → SrCrO\(_4\) The reaction shows a 1:1 stoichiometry between the Sr\(^{2+}\) ions and the CrO\(_4^{2-}\) ions. Therefore, for every mole of Sr\(^{2+}\) ion, we need one mole of CrO\(_4^{2-}\) ion. Moles of Na\(_2\)CrO\(_4\) required = moles of Sr\(^{2+}\) ions = 0.00644 mol

Calculate the volume of Na\(_2\)CrO\(_4\) solution needed

Finally, we will use the concentration of Na\(_2\)CrO\(_4\) solution to find the volume required to deliver the needed moles of Na\(_2\)CrO\(_4\): Volume of Na\(_2\)CrO\(_4\) solution = moles / concentration = 0.00644 mol / 0.0245 M = 0.2629 L So, 0.2629 L or 262.9 mL of the Na\(_2\)CrO\(_4\) solution is needed to precipitate all the Sr\(^{2+}\)(aq) as SrCrO\(_4\).

Key concepts

Molar Mass Calculation

Before diving into calculating molar masses, let's understand what it is. The molar mass of a substance is the mass of one mole of its entities, be it atoms, molecules, or ions. It is expressed in grams per mole (g/mol). To calculate the molar mass, you sum up the atomic masses of all atoms present in the chemical formula.
For example, in the compound Sr(NOdots_3)_2, the atomic masses are as follows: Sr is 87.62 g/mol, nitrogen (N) is 14.01 g/mol, and oxygen (O) is 16.00 g/mol. Therefore, the molar mass of Sr(NOdots_3)_2 is \[1 \times 87.62 + 2 \times (1 \times 14.01 + 3 \times 16.00) = 211.63 \text{ g/mol}.\]This value is essential for converting grams to moles and vice versa, along with determining concentrations in solutions.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They are represented by chemical equations that show the exact molar relationships between the substances involved. During a reaction, bonds are broken and new bonds are formed, resulting in different substances.
Let's take a closer look at the precipitation reaction in the exercise: \[\text{Sr}^{2+} + \text{CrO}_4^{2-} \rightarrow \text{SrCrO}_4. \]Here, strontium ions (\text{Sr}^{2+}) react with chromate ions (\text{CrO}_4^{2-}) to form strontium chromate (\text{SrCrO}_4), a solid precipitate that settles out of the solution. The equation represents a 1:1 ratio, meaning for every mole of Sr^{2+}, one mole of CrO_4^{2-} is required. Understanding stoichiometry is crucial here as it helps in calculating how much of a reactant is needed or how much product will form.

Solution Concentration

Solution concentration is a measure of the amount of solute dissolved in a solvent. It is often represented in molarity (M), which is moles of solute per liter of solution. Understanding concentration helps in preparing solutions of a specific strength and for predicting how substances will react together.
In our example, the concentration of the Sr(NO\dots_3)_2 solution is determined by \[\text{Concentration} = \frac{\text{Moles of solute}}{\text{Volume of solution}}.\] Here, concentration = 0.0644 M meaning each liter of the solution has 0.0644 moles of Sr(NO\dots_3)_2. Further calculations show us how much of the substance is in a smaller sample of the solution, crucial when preparing titrations or predicting outcomes of reactions.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid known as a precipitate. They are a subtype of double displacement reactions, where the cations and anions of the reactants exchange partners.
In this exercise, the precipitation reaction is between Sr^{2+}(aq) ions and CrO_4^{2-}(aq) ions to form SrCrO_4(s), which precipitates out of the solution. The reaction shows a 1:1 stoichiometry, essential for predicting the quantities of reactants needed to entirely precipitate the desired product. Why does precipitation happen? It occurs because the product is insoluble in water, unlike the reactants that remain dissolved. Understanding when and how precipitation occurs is crucial for many applications in industrial processes, water treatment, and laboratory experiments.