Question

The mass percentage of chloride ion in a \(25.00-\mathrm{mL}\) sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took \(42.58 \mathrm{~mL}\) of \(0.2997 \mathrm{M}\) silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in the seawater if its density is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

Short answer

The mass percentage of chloride ions in the seawater can be calculated using the following steps: 1. Calculate the moles of silver nitrate used in titration: Moles of silver nitrate = \(0.2997 \mathrm{M} \times 0.04258 \mathrm{L} = 0.0127593 \, \text{mol} \). 2. Determine the moles of chloride ions, which is equal to the moles of silver nitrate: Moles of chloride ions = \(0.0127593 \, \text{mol} \). 3. Calculate the mass of chloride ions: Mass of chloride ions = \(0.0127593 \, \text{mol} \times 35.45 \frac{\mathrm{g}}{\mathrm{mol}} = 0.4524 \,\mathrm{g} \). 4. Calculate the mass of the seawater sample: Mass of seawater = \(25.00 \,\mathrm{mL} \times 1.025 \frac{\mathrm{g}}{\mathrm{mL}} = 25.625 \,\mathrm{g} \). 5. Calculate the mass percentage of chloride ions in the seawater: Mass percentage of chloride ions = \(\frac{0.4524 \,\mathrm{g}}{25.625 \,\mathrm{g}} \times 100 = 1.766 \% \). Therefore, the mass percentage of chloride ions in the seawater is approximately \(1.766 \% \).

Step-by-step solution

Calculate the moles of silver nitrate used in the titration

Since we know the volume and molarity of silver nitrate used, we can find the moles of silver nitrate using the formula: Moles of silver nitrate = Molarity × Volume Remember to convert the volume to liters before plugging the values into the formula.

Determine the moles of chloride ions

Since both silver nitrate and chloride ions form a 1:1 ratio in the reaction, the moles of chloride ions are equal to the moles of silver nitrate. Thus, use the moles of silver nitrate calculated in Step 1 as the moles of chloride ions.

Calculate the mass of chloride ions

Now that we know the number of moles of chloride ions, we can determine the mass of chloride ions using the molar mass of chloride (35.45 g/mol). The mass can be calculated using the formula: Mass of chloride ions = Moles of chloride ions × Molar mass of chloride

Calculate the mass of the seawater sample

Using the volume and density of the seawater sample, we can find the mass of the seawater using the formula: Mass of seawater = Volume × Density

Calculate the mass percentage of chloride ions in the seawater

Finally, we can determine the mass percentage of chloride ions in the seawater. The mass percentage can be calculated using the following formula: Mass percentage of chloride ions = (Mass of chloride ions / Mass of seawater) × 100 Plug in the values determined in steps 3 and 4, and calculate the mass percentage.

Key concepts

Titration

Titration is a laboratory technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration, called the titrant. In our exercise, the unknown solution is seawater containing chloride ions, and the titrant is a silver nitrate solution. The process of titration involves adding the titrant to the sample until the chemical reaction is complete, which is indicated by the equivalence point. This point can be detected by a color change if an appropriate indicator is used, or by the fact that no more precipitate forms.

During the titration, the silver ions from silver nitrate react with chloride ions, forming a solid precipitate of silver chloride. By exactly measuring how much titrant is needed to fully react with all chloride ions present, we can calculate the chloride ion concentration in the seawater.

Molarity

Molarity is a measure of concentration that expresses the number of moles of a solute contained in one liter of solution. The molarity formula is represented by:
\[\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\]

The molarity of the silver nitrate (\(0.2997 \mathrm{M}\)) in the given problem indicated the strength of the titrant. Knowing the molarity allows us to determine the number of moles of the titrant used in the titration, which is critical for finding the moles of the chloride ions present in the seawater sample.

Equivalence Point

The equivalence point of a titration is the precise point at which the amount of titrant added is stoichiometrically equivalent to the quantity of substance present in the sample. In our exercise, the equivalence point occurs when all chloride ions \(\text{Cl}^-\) in the seawater have reacted with the silver ions \(\text{Ag}^+\) to form silver chloride (\(\text{AgCl}\)). At the equivalence point, the moles of \(\text{Ag}^+\) added from the silver nitrate solution are equal to the moles of \(\text{Cl}^-\) in the sample. The volume of silver nitrate solution used to reach the equivalence point provides the information needed to calculate the mass percentage of chloride ion in the seawater.

Precipitation Reaction

Precipitation reactions occur when two soluble salts react in solution to form an insoluble product, known as a precipitate. In the given problem, when silver nitrate (\(\text{AgNO}_3\)) reacts with the chloride ions in seawater, an insoluble salt—silver chloride (\(\text{AgCl}\))—is formed. This reaction can be represented by the chemical equation:

\[\text{Ag}^+ (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s)\]

The silver chloride precipitates out of the solution, and its formation is essential in determining when the equivalence point of the titration has been reached. Because silver chloride is not soluble in water, it forms as a solid, allowing it to be separated from the remaining aqueous mixture.

Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It involves calculations based on the balanced chemical equations to determine the amount of reactants needed or products formed. In the context of our exercise, stoichiometry allows us to connect the moles of silver nitrate used with the moles of chloride ions present. Because the balanced chemical equation for the reaction of silver nitrate and chloride ions shows a 1:1 mole ratio, we know that each mole of silver nitrate reacts with one mole of chloride ions. Thus, stoichiometry is crucial for calculating the moles, and consequently the mass, of chloride ions in the seawater, which then leads to determining the mass percentage of chloride ion in the sample.