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: \(\operatorname{CuSO}_{4}(a q)+\mathrm{Na}_{2} S(a q) \rightarrow \operatorname{CuS}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) What volume of \(0.105 M \mathrm{Na}_{2} \mathrm{S}\) 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 CuSO4 solution, 31.67 mL of 0.105 M Na2S solution will be required.

Step-by-step solution

Write down the given information

We have been given: - Copper(II) sulfate solution: 27.5 mL and 0.121 M - Sodium sulfide solution: 0.105 M - Balanced equation: \(\operatorname{CuSO}_{4}(a q) + \mathrm{Na}_{2} S(a q) \rightarrow \operatorname{CuS}(s) + \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\)

Calculate the moles of CuSO4

First, we need to find the moles of CuSO4 using the given concentration and volume: Moles of CuSO4 = concentration × volume Moles of CuSO4 = 0.121 M × 0.0275 L = 0.003325 mol

Determine the stoichiometric ratio of CuSO4 to Na2S

According to the balanced equation, one mole of CuSO4 reacts with one mole of Na2S: \(\operatorname{CuSO}_{4}(a q) + \mathrm{Na}_{2} S(a q) \rightarrow \operatorname{CuS}(s) + \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) 1 mol CuSO4 = 1 mol Na2S Since the moles of CuSO4 = 0.003325 mol, the moles of Na2S required to react completely is also 0.003325 mol.

Calculate the volume of the Na2S solution

Now we need to find the volume of the Na2S solution to produce the required moles of Na2S using the given concentration: Volume of Na2S solution = moles of Na2S / concentration of Na2S Volume of Na2S solution = 0.003325 mol / 0.105 M = 0.0316667 L = 31.67 mL So, it would require 31.67 mL of 0.105 M Na2S solution to precipitate all the copper(II) ion from the 27.5 mL of 0.121 M CuSO4 solution.

Key concepts

Chemical Reactions

When substances interact to form new compounds, a chemical reaction occurs. In our context, the reaction is between copper(II) sulfate and sodium sulfide. This process is commonly depicted using a chemical equation that represents the reactants and products:

• **Reactants:** Substances that start the reaction. Here, these are copper(II) sulfate (\(\text{CuSO}_4\)) and sodium sulfide (\(\text{Na}_2\text{S}\)).
• **Products:** New substances formed as a result of a reaction. In this case, these are copper sulfide (\(\text{CuS}\)) and sodium sulfate (\(\text{Na}_2\text{SO}_4\)).

Understanding chemical reactions is essential because they allow us to predict how different substances will behave when mixed. Balancing these equations is important to ensure the conservation of atoms, which reflects the unchangeable nature of matter.

Precipitation

Precipitation in chemistry refers to the formation of a solid from a solution. This often occurs when two dissolved compounds are mixed and react to form an insoluble product. In this example, the reaction yields copper sulfide (\(\text{CuS}\)), which is a solid that precipitates out of the solution.

• **Solubility:** A measure of how well a substance dissolves in a solvent. Copper sulfide's low solubility causes it to precipitate.
• **Driving Force:** The precipitation itself acts as a driving force by removing ions from the solution, which supports the forward reaction.

Precipitation reactions are important in various applications like water purification, where undesired metals are removed from solutions by converting them to insoluble solids.

Solution Concentration

Solution concentration describes how much solute is dissolved in a given amount of solvent. It is usually expressed in molarity (M), which is moles of solute per liter of solution. In this problem:

• **Copper(II) sulfate concentration:** 0.121 M means that there are 0.121 moles of copper(II) sulfate in one liter.
• **Sodium sulfide concentration:** 0.105 M represents 0.105 moles of sodium sulfide per liter.

Calculating concentrations helps us determine how substances will react and interact. Having accurate concentrations is crucial for predicting reaction yields and for industrial applications where precise formulations are essential.

Moles Calculation

Calculating moles in a solution is vital for stoichiometric calculations in chemistry. Here, the process involves converting the volume and concentration of a solution to moles, which allows chemists to predict how much of each reactant is needed or produced in a reaction.For example:

• **Copper(II) sulfate:** Using the formula \(\text{Moles} = \text{Concentration} \times \text{Volume}\), we calculate the moles as \(0.121 \times 0.0275 = 0.003325\) moles.
• **Sodium sulfide:** Similarly, we can find the moles needed by applying the stoichiometric ratio from the balanced equation.

These calculations form the backbone of chemical experimentation, allowing precise manipulation of reactants to achieve desired results. Understanding how to convert between volumes, concentrations, and moles is an essential skill in chemistry.