Question

One way to determine the amount of chloride ion in a water sample is to titrate the sample with standard \(\mathrm{AgNO}_{3}\) solution to produce solid AgCl. $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s)$$ If a 25.0 -mL water sample requires \(27.2 \mathrm{mL}\) of \(0.104 M \mathrm{AgNO}_{3}\) in such a titration, what is the concentration of \(\mathrm{Cl}^{-}\) in the sample?

Short answer

The concentration of \(\mathrm{Cl^{-}}\) in the water sample is approximately \(0.114\mathrm{M}\).

Step-by-step solution

Write down the given information

We are given the following information: - Volume of the water sample: \(V_{2}=25.0 \mathrm{mL}\) - Volume of the \(\mathrm{AgNO_{3}}\) solution used: \(V_{1}=27.2 \mathrm{mL}\) - Concentration of the \(\mathrm{AgNO_{3}}\) solution: \(C_{1}=0.104 \mathrm{M}\) We know that the stoichiometry of the reaction is 1:1, which means that 1 mole of \(\mathrm{Ag^{+}}\) reacts with 1 mole of \(\mathrm{Cl^{-}}\). Our goal is to find the concentration of Cl⁻ in the water sample: \(C_{2}\).

Use the titration formula to find the concentration of Cl⁻

We can use the titration formula to solve for \(C_{2}\): $$C_{1}V_{1}=C_{2}V_{2}$$ Rearrange the formula to solve for \(C_{2}\), $$C_{2}=\frac{C_{1}V_{1}}{V_{2}}$$ Now, plug in the known values: $$C_{2}=\frac{(0.104 \mathrm{M})(27.2 \mathrm{mL})}{25.0 \mathrm{mL}}$$

Calculate the concentration of Cl⁻

Calculate the value for \(C_{2}\): $$C_{2}= \frac{(0.104 \times 27.2)}{25.0}$$ $$C_{2} \approx 0.114 \mathrm{M}$$ Therefore, the concentration of \(\mathrm{Cl^{-}}\) in the water sample is approximately \(0.114\mathrm{M}\).

Key concepts

Stoichiometry

Stoichiometry plays a crucial role in understanding chemical reactions and their quantitative aspects. It involves the calculations of reactants and products in a chemical reaction. One key part of stoichiometry is understanding mole ratios from balanced chemical equations. In the example provided, the chemical reaction \[\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq) \rightarrow \mathrm{AgCl}(s)\] shows a 1:1 stoichiometry. This means one mole of silver ions \(\mathrm{Ag}^{+}\) reacts with one mole of chloride ions \(\mathrm{Cl}^{-}\) to form one mole of silver chloride \(\mathrm{AgCl}\).

This simple mole-to-mole relationship allows us to use the amounts of \(\mathrm{AgNO}_{3}\) titrated to calculate the chloride concentration in the water sample. By recognizing this stoichiometry, we know exactly how the ions are reacting and can rely on these ratios to find missing concentrations with confidence.

Chemical Reaction

Understanding the chemical reactions underlying a titration is essential to grasp the full picture. A titration is a method used to determine the concentration of a substance in a solution by adding a reagent of known concentration and volume until the reaction reaches its endpoint or completion.

In the given exercise, the chemical reaction of concern is between silver nitrate \(\mathrm{AgNO}_{3}\) and chloride ions \(\mathrm{Cl}^{-}\) from the sample to form silver chloride \(\mathrm{AgCl}\) as a solid precipitate. The chemical equation \[\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq) \rightarrow \mathrm{AgCl}(s)\] characterizes this process. It's crucial to note the physical state of the products and reactants. In aqueous conditions (denoted by \((aq)\)), the ions are dissolved in water and free to react. The formation of \(\mathrm{AgCl}(s)\) indicates the solid precipitate formation which signals the completion of the titration.

Concentration Calculation

Concentration calculation is a vital competency in chemistry, especially when dealing with solutions and reactions like titrations. The concentration of a solution is often expressed in molarity (M), which is defined as moles of solute per liter of solution.

In the exercise, we saw how to calculate the concentration of chloride ions in the water sample after titration. The relationship used here follows \[C_{1}V_{1} = C_{2}V_{2}\] where \(C_{1}\) and \(V_{1}\) are the concentration and volume of the \(\mathrm{AgNO}_{3}\) used, and \(C_{2}\) and \(V_{2}\) are the concentration and volume of the chloride ions in the sample.

Rearranging to solve for \(C_{2}\) gives:\[C_{2} = \frac{C_{1}V_{1}}{V_{2}}\] By plugging in the values, we calculate the concentration \(C_{2}\) as approximately \(0.114\, \mathrm{M}\). This equation highlights how balance in stoichiometry and reaction completeness are pivotal in deriving solution concentrations from titration data.