Question

You want to analyze a cadmium nitrate solution. What mass of \(\mathrm{NaOH}\) is needed to precipitate the \(\mathrm{Cd}^{2+}\) ions from \(35.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) solution?

Short answer

To precipitate the Cd^(2+) ions from 35.0 mL of 0.500 M Cd(NO₃)₂ solution, 1.40 grams of NaOH is required.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between NaOH and Cd(NO₃)₂ is: \[Cd(NO_3)_2(aq) + 2NaOH(aq) \rightarrow Cd(OH)_2(s) + 2NaNO_3(aq)\] In this equation, one mole of Cd(NO₃)₂ reacts with two moles of NaOH to produce one mole of cadmium hydroxide (Cd(OH)₂) precipitate and two moles of sodium nitrate (NaNO₃).

Calculate moles of Cd(NO₃)₂ in the solution

We can calculate the moles of Cd(NO₃)₂ using the given volume and concentration: Moles of Cd^(2+) ions = Volume (L) × Concentration (M) First, convert the volume of the solution from milliliters to liters (L): \[V = 35.0\, mL = 35.0 \times 10^{-3}\, L = 0.035\, L\] Now, calculate the moles of Cd^(2+) ions: \[Moles\,of\,Cd(NO_3)_2 = 0.035 L \times 0.500 M = 0.0175\,mol\]

Determine the moles of NaOH required

From the balanced chemical equation, we know that 1 mole of Cd(NO₃)₂ requires 2 moles of NaOH to completely react: Moles of NaOH = Moles of Cd(NO₃)₂ × 2 \[Moles\,of\,NaOH = 0.0175\,mol \times 2 = 0.035\,mol\]

Calculate the mass of NaOH needed

Finally, calculate the mass of NaOH using its molar mass (40.0 g/mol): Mass of NaOH (g) = Moles of NaOH × Molar mass of NaOH (g/mol) \[Mass\,of\,NaOH = 0.035\,mol \times 40.0\,g/mol = 1.40\,g\] Therefore, 1.40 grams of NaOH is required to precipitate the Cd^(2+) ions from 35.0 mL of 0.500 M Cd(NO₃)₂ solution.

Key concepts

Chemical Equations

Chemical equations are used to represent chemical reactions. They provide a clear and organized way to show the substances involved in the reaction and how they transform. In a balanced chemical equation, the reactants (the starting substances) and products (the resulting substances) have their atoms accounted for, ensuring mass conservation. For example, in the reaction between cadmium nitrate (Cd(NO_3)_2) and sodium hydroxide (NaOH), the balanced equation is:\[Cd(NO_3)_2(aq) + 2NaOH(aq) \rightarrow Cd(OH)_2(s) + 2NaNO_3(aq)\]This equation tells us a lot:

• One mole of cadmium nitrate reacts with two moles of sodium hydroxide.
• The product is one mole of cadmium hydroxide and two moles of sodium nitrate.
• The state of matter for each compound: aqueous (aq) and solid (s).

Balancing chemical equations is crucial because it reflects the principle of conservation of mass, ensuring that the same number of each type of atom appears on both sides of the equation. It also helps us to understand the stoichiometry of the reaction, determining the proportions of reactants and products needed.

Moles Calculation

Calculating moles is a foundational skill in stoichiometry. It involves determining the amount of a substance, in moles, present in a given volume and concentration or as related to another substance. When dealing with solutions, the formula for calculating moles is:\[\text{Moles} = \text{Volume (L)} \times \text{Concentration (M)}\]In the given problem, we determine the moles of cadmium nitrate in the solution using its volume (converted to liters) and its molarity:

• Volume in liters: 35.0 mL = 0.035 L
• Concentration: 0.500 M

Therefore, the moles of cadmium nitrate are:\[\text{Moles of } Cd(NO_3)_2 = 0.035 \text{ L} \times 0.500 \text{ M} = 0.0175 \text{ mol}\]This calculation allows us to understand how much of the compound is available to participate in the chemical reaction. With the balanced equation, we can determine the amount of another reactant needed or the amount of product formed, based on the stoichiometric relationships shown.

Precipitation Reaction

A precipitation reaction involves the formation of an insoluble solid from two soluble substances reacting in a solution. When cadmium nitrate reacts with sodium hydroxide, a precipitation reaction occurs, resulting in the formation of cadmium hydroxide as a solid:\[Cd(OH)_2 ext{ (s)}\]This solid forms because it's insoluble in water, and it separates from the solution. Identifying such reactions can be crucial in processes like purification, separation, or even analyzing different salts and metals.The formation of a precipitate can tell us about the solubility rules. For instance, metal hydroxide precipitates are somewhat common, especially with transition metals like cadmium. Solubility rules help predict whether a reaction forms a precipitate:

• Most hydroxides are insoluble except those of sodium, potassium, and ammonium.
• Many salts containing transition metals form precipitates with hydroxides.

Understanding precipitation reactions is key in areas such as qualitative analysis and separating components from complex mixtures. It allows for the designed crystallization of components, further leveraging their physical properties.