Question

A \(450.0-\mathrm{mL}\) sample of a \(0.257-M\) solution of silver nitrate is mixed with \(400.0 \mathrm{~mL}\) of \(0.200 M\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

Short answer

After the reaction is complete, the concentration of Cl⁻ ions in the solution is 0.052 M.

Step-by-step solution

Write the balanced chemical equation for the reaction

First, we need to write the balanced chemical equation to describe the reaction between silver nitrate (AgNO₃) and calcium chloride (CaCl₂): \(2\mathrm{AgNO_3} + \mathrm{CaCl_2} \rightarrow \mathrm{2AgCl} + \mathrm{Ca(NO_3)_2}\)

Calculate the initial moles of both reactants

Next, we need to find the initial moles of both reactants. We can use the initial volume and concentration of each solution to do this: - moles of AgNO₃ = volume (L) × concentration (M) - moles of CaCl₂ = volume (L) × concentration (M) Moles of AgNO₃ = \( 450.0 \mathrm{~mL} * \frac {1\mathrm{~L}}{1000\mathrm{~mL}} * 0.257 \mathrm{~M} = 0.11565\ \mathrm{mol}\) Moles of CaCl₂ = \( 400.0\mathrm{~mL} * \frac {1\mathrm{~L}}{1000\mathrm{~mL}} * 0.200\ \mathrm{M} = 0.08000\ \mathrm{mol}\)

Determine the limiting reagent and reactants' moles that remain after the reaction

Using the balanced chemical equation, we can compare the stoichiometric ratios of the two reactants: Stoichiometric ratio = \( \frac{2\mathrm{mol\ AgNO_3}}{1\mathrm{mol\ CaCl_2}}\) Now we can determine which reactant is the limiting reagent by calculating the stoichiometric amount of the other reactant for each reactant: - Amount of CaCl₂ needed for 0.11565 mol of AgNO₃: \( \frac{0.11565\ \mathrm{mol\ AgNO_3}}{2} = 0.057825\ \mathrm{mol\ CaCl_2} \) - Amount of AgNO₃ needed for 0.08000 mol of CaCl₂: \( 0.08000\ \mathrm{mol\ CaCl_2} * 2 = 0.16000\ \mathrm{mol\ AgNO_3} \) Since we have enough CaCl₂ for all the available AgNO₃ to react, but insufficient AgNO₃ to react with all of the CaCl₂, AgNO₃ is the limiting reagent. Thus, only 0.057825 mol of CaCl₂ reacts with the available AgNO₃, leaving 0.022175 mol of CaCl₂ (0.08000 - 0.057825) that did not react.

Calculate the total volume of the solution after mixing

The final volume of the solution will be the sum of the initial volumes of the two solutions: Total volume = initial volume of AgNO₃ solution + initial volume of CaCl₂ solution Total volume = 450.0 mL + 400.0 mL = 850.0 mL = 0.850 L

Calculate the final concentration of Cl⁻ ions in the solution after the reaction

Now, we can calculate the final concentration of Cl⁻ in the solution by dividing the moles of Cl⁻ remaining by the final volume of the solution: Concentration of Cl⁻ = \( \frac{0.022175\ \mathrm{mol\ CaCl_2} * 2 \mathrm{mol\ Cl^-}}{0.850\ \mathrm{L}} = 0.052 \mathrm{~M}\) This result means that after the reaction is complete, the concentration of Cl⁻ ions in the solution is 0.052 M.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the substances involved in chemical reactions. It relies on the basic principle that matter is conserved during a reaction, meaning the amount of reactants used and products formed is predictable.

For students tackling stoichiometry, a balanced chemical equation is essential as it provides the ratios in which chemicals react and form products. To understand how much of one substance reacts with another, we convert the volumes and concentrations given (commonly in molarity, which is moles per liter) to moles. The stoichiometric coefficients from the balanced equation tell us the exact proportions needed to react completely without any reactant left over.

For instance, in our exercise, when working out how much silver nitrate reacts with calcium chloride, the balanced equation shows a 2:1 ratio. If these ratios are confusing, consider using visuals, such as a Punnett square, to make the concept clearer.

Limiting Reagent

The limiting reagent in a chemical reaction is the substance that is completely consumed first and thus determines when the reaction stops. There will be no additional product formed once this reagent is exhausted, even if other reactants are still present in excess.

To find the limiting reagent, compare the amount of each reactant that is actually present to the amount that would be needed to completely react with the other substances, based on the stoichiometric proportions. A tip to aid understanding is to imagine the reactants as guests and the product as cakes at a party. The limiting reagent is like the guest who eats more quickly and determines when all the cake is gone. In our exercise, the calculation of moles of both substances and comparison through the stoichiometric ratio allowed us to identify AgNO₃ as the limiting reagent.

Moles and Molarity

Understanding moles and molarity is crucial when calculating concentrations of solutions. A mole is the SI unit for amount of substance, representing approximately 6.022 x 10²³ particles (Avogadro's number). Molarity, on the other hand, is a measure of concentration, given in moles per liter.

When working with solutions, students have to be adept at converting between volume and moles using molarity. Remember, to calculate moles from volume and molarity, use the formula: moles = volume (L) × molarity (M). Services like online calculators or molarity charts can be helpful tools for practicing converting between these units. In the example problem, using molarity and volume, we found the initial number of moles for the reactants before the reaction.

Chemical Equation Balancing

Balancing a chemical equation is a fundamental step in chemistry. It ensures that the number of atoms for each element is the same on both sides of the equation, respecting the Law of Conservation of Mass.

To balance an equation, begin by writing the skeleton equation with correct formulas for reactants and products. Then adjust coefficients to get the same number of each atom on both sides. Sometimes using a systematic approach, like balancing elements that appear in only one reactant and one product first and leaving more common elements for last, simplifies the process. In the example, the chemical equation for the reaction between silver nitrate and calcium chloride was balanced by adjusting coefficients to two for AgNO₃ and one for CaCl₂, following the stoichiometry of the reaction.