Question

Many metal ions are precipitated from solution by the sulfide ion. As an example, consider treating a solution of copper(II) sulfate with sodium sulfide solution: $$ \mathrm{CuSO}_{4}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CuS}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) $$ What volume of \(0.105 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}\) solution would be required to precipitate all of the copper(II) ion from \(27.5 \mathrm{~mL}\) of \(0.121 \mathrm{M} \mathrm{CuSO}_{4}\) solution?

Short answer

To precipitate all the copper(II) ions from \(27.5 mL\) of \(0.121 M\) copper(II) sulfate solution, approximately \(31.69 mL\) of \(0.105 M\) sodium sulfide solution is required.

Step-by-step solution

Write down the given information

We are given the following: - Volume of copper(II) sulfate solution: \(27.5 mL\) - Concentration of copper(II) sulfate solution: \(0.121 M\) - Concentration of sodium sulfide solution: \(0.105 M\) Our goal is to determine the volume of sodium sulfide solution needed to precipitate all of the copper(II) ions from the copper(II) sulfate solution.

Calculate the moles of copper(II) sulfate

Using the given volume and concentration of copper(II) sulfate solution, we can calculate the moles of copper(II) sulfate. Moles of copper(II) sulfate = Molarity × Volume Moles of CuSO4 = (0.121 M) × (27.5 mL) × \(\frac{1 L}{1000 mL}\) Moles of CuSO4 = 0.0033275 mol

Determine the stoichiometric ratio between copper(II) sulfate and sodium sulfide

From the balanced chemical equation, we have: $$ \mathrm{CuSO}_{4}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CuS}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) $$ The stoichiometric ratio between copper(II) sulfate (CuSO4) and sodium sulfide (Na2S) is 1:1, i.e., 1 mole of CuSO4 reacts with 1 mole of Na2S.

Calculate the moles of sodium sulfide needed

Now, we can calculate the moles of sodium sulfide needed, based on the stoichiometric ratio. Since the ratio is 1:1, the moles of sodium sulfide needed are equal to the moles of copper(II) sulfate. Moles of Na2S needed = 0.0033275 mol

Calculate the volume of sodium sulfide solution required

Finally, we can calculate the volume of sodium sulfide solution required using the moles of Na2S needed and the given concentration of the Na2S solution. Volume of Na2S solution = moles of Na2S / Molarity of Na2S solution Volume of Na2S solution = 0.0033275 mol / 0.105 M Volume of Na2S solution = 0.03169047 L Since we want the volume in milliliters, we need to convert the volume to milliliters. Volume of Na2S solution = 0.03169047 L × \(\frac{1000 mL}{1 L}\) = 31.69 mL Therefore, the volume of 0.105 M sodium sulfide solution required to precipitate all of the copper(II) ions from 27.5 mL of 0.121 M copper(II) sulfate solution is approximately 31.69 mL.

Key concepts

Chemical Precipitation

Chemical precipitation is a fundamental concept in chemistry where a solid forms in a solution during a chemical reaction. This solid is known as the precipitate, and the process is often used to separate substances and purify solutions. For instance, in our exercise, copper sulfide (CuS) forms as a solid when a solution of copper(II) sulfate (CuSO4) reacts with sodium sulfide (Na2S).

Understanding how and why certain compounds precipitate is crucial for various applications, including water treatment, where undesired ions are removed by precipitating them out of the water. Precipitation reactions often obey the solubility rules, which predict the formation of solids based on the nature of the ions involved. In the given exercise, copper sulfide precipitates because it is not soluble in water, unlike the sodium sulfate (Na2SO4) that remains in the aqueous phase.

Molar Concentration

Molar concentration, often represented by the unit M (molarity), refers to the number of moles of a substance per liter of solution. It is a key concept when discussing solutions and their reactivity in chemical processes. Knowing the molarity of a solution allows chemists to calculate the number of moles present in a given volume of solution, which is essential for quantitative analysis and reaction predictions.

The molar concentration gives us a clear understanding of the amount of reactant available to participate in a reaction, as seen in our textbook problem where the concentrations of CuSO4 and Na2S solutions were provided. To find out how the solution's molarity relates to the chemical precipitation process, students must be able to efficiently convert between moles, volume, and molarity.

Stoichiometric Calculations

Stoichiometric calculations are vital in chemistry, as they allow chemists to predict the outcomes of reactions and to determine the right proportions of reactants and products. They are based on the stoichiometry of a chemical reaction, which is the quantitative relationship between reactants and products represented in a balanced chemical equation.

These calculations often involve determining the amounts of reactants needed or products formed in a chemical reaction. In the exercise, for example, we used stoichiometric calculations to find out how much sodium sulfide is required to completely react with the copper(II) sulfate. The balanced chemical equation showed a 1:1 mole ratio between CuSO4 and Na2S, which is the key to stoichiometric calculations. Whether you're scaling up a reaction for industrial production or simply completing a lab assignment, understanding stoichiometry is indispensable for any chemist.