Question

The amount of nickel(II) present in an aqueous solution can be determined by precipitating the nickel with the organic chemical reagent dimethylglyoxime [ \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{NOH}) \mathrm{C}(\mathrm{NOH}) \mathrm{CH}_{3}\) commonly abbreviated as "DMG"]. \(\mathrm{Ni}^{2+}(a q)+2 \mathrm{DMG}(a q) \rightarrow \mathrm{Ni}(\mathrm{DMG})_{2}(s)\) How many milliliters of \(0.0703 \mathrm{M}\) DMG solution is required to precipitate all the nickel(II) present in \(10.0 \mathrm{~mL}\) of \(0.103 \mathrm{M}\) nickel \((\mathrm{II})\) sulfate solution?

Short answer

Approximately 29.3 mL of 0.0703 M DMG solution is required to precipitate all the nickel(II) present in 10.0 mL of 0.103 M nickel(II) sulfate solution.

Step-by-step solution

Write down the given information

We are given the following information: - Volume of nickel(II) sulfate solution: 10.0 mL - Concentration of nickel(II) sulfate solution: 0.103 M - Concentration of DMG solution: 0.0703 M We need to find the volume of DMG solution required to precipitate all the nickel(II) ions.

Calculate the moles of nickel(II) ions

We can calculate the moles of nickel(II) ions by multiplying the concentration and the volume of the nickel(II) sulfate solution. Moles of nickel(II) ions = concentration × volume (in liters) Moles of nickel(II) ions = 0.103 M × (10.0 mL/1000) Moles of nickel(II) ions = 0.00103 mol

Calculate the moles of DMG required

Using the stoichiometry of the reaction, we can calculate the moles of DMG required to precipitate all the nickel(II) ions. According to the reaction equation, 1 mol of Ni²⁺ ions react with 2 moles of DMG: Ni²⁺(aq) + 2 DMG(aq) → Ni(DMG)₂(s) Moles of DMG required = 2 × moles of nickel(II) ions Moles of DMG required = 2 × 0.00103 mol Moles of DMG required = 0.00206 mol

Calculate the volume of DMG solution required

Now we can calculate the required volume of the DMG solution by dividing the moles of DMG required by its concentration. Volume of DMG solution (in liters) = moles of DMG required / concentration of DMG solution Volume of DMG solution (in liters) = 0.00206 mol / 0.0703 M Volume of DMG solution (in liters) = 0.0293 L Convert the volume from liters to milliliters: Volume of DMG solution (in mL) = 0.0293 L × 1000 Volume of DMG solution (in mL) = 29.3 mL So, approximately 29.3 mL of 0.0703 M DMG solution is required to precipitate all the nickel(II) present in 10.0 mL of 0.103 M nickel(II) sulfate solution.

Key concepts

Stoichiometry Calculation

Understanding stoichiometry calculations is essential when dealing with chemical reactions, as it allows us to predict amounts of reactants consumed and products formed. In the case of our nickel(II) and DMG reaction, stoichiometry describes the quantitative relationship between the reactants and the products.

Given the balanced chemical equation, \(\mathrm{Ni}^{2+}(aq) + 2 \mathrm{DMG}(a q) \rightarrow \mathrm{Ni}(\mathrm{DMG})_{2}(s)\), the ratio of moles of nickel(II) ions to moles of DMG is 1:2. This stoichiometric ratio is crucial as it forms the basis for our calculations. We determine the amount of one reactant needed to completely react with a given amount of the other. In practice, perform stoichiometry calculations by:

• Identifying the number of moles of a known reactant or product.
• Using the stoichiometric ratio from the balanced equation to find the moles of the unknown substance.
• Converting these moles into the desired units, such as volume or mass.

These steps provide a clear and concise path to finding how much of one reactant is necessary to react fully with a given amount of the other reactant, which in our exercise is the volume of DMG solution needed to react with the nickel(II) ions.

Mole Concept

The mole concept is a fundamental principle of chemistry that relates the mass of a substance to its number of entities, such as atoms, ions, or molecules. One mole is defined as exactly \(6.022 \times 10^{23}\) entities, which is Avogadro's number. In stoichiometric calculations, the mole concept serves as a bridge between the macroscopic world of grams and liters and the microscopic world of molecules and atoms.

In our example with the nickel(II) precipitation, understanding the mole concept allows us to convert the given concentrations and volumes into moles, the key step in performing stoichiometric calculations. To illustrate, the volume of nickel(II) sulfate solution, \(10.0 \mathrm{~mL}\), and its concentration, \(0.103 \mathrm{M}\), are used to calculate the moles of nickel(II) ions present. This conversion is vital because stoichiometry deals with moles rather than direct volumes or masses. Therefore, grasping the mole concept is pivotal in solving quantitative problems in chemistry.

Chemical Precipitation

Chemical precipitation is a process where a solid, known as a precipitate, forms from a solution during a chemical reaction. This occurs when the product of the reaction has low solubility in the solution, causing it to separate out as a solid. Precipitation is often utilized in analytical chemistry to isolate and quantify specific ions, like nickel(II) ions in our exercise.

The precipitation of nickel(II) ions with DMG is a highly specific and sensitive reaction, common in qualitative and quantitative analyses of nickel. These precipitation reactions are influenced by factors such as the concentrations of reactants, the solubility of the products, and the temperature of the solution. In stoichiometry, the concept of chemical precipitation helps us to understand the exact amounts of reactants needed to form a precipitate and ensure that the reaction goes to completion; hence no nickel(II) ions remain in the solution. By carefully calculating the stoichiometry, we can use precipitation to not only identify the presence of certain ions but also quantify their amount with precision.