Problem 80 - Chemistry The Central Science

Question

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?

Step-by-Step Solution

Answer

The mass of $\mathrm{NaOH}$ needed to precipitate the $\mathrm{Cd}^{2+}$ ions from $35.0 \mathrm{~mL}$ of $0.500 \mathrm{M} \mathrm{Cd(NO_3)_2}$ solution is approximately 1.40 grams.

Step 1: Write the balanced chemical equation

The balanced equation for the reaction between sodium hydroxide and cadmium nitrate is: \( \mathrm{Cd(NO_3)_2 + 2NaOH \rightarrow Cd(OH)_2 + 2NaNO_3} \) This equation tells us that 2 moles of NaOH are required to precipitate 1 mole of Cd^2+ ions as cadmium hydroxide (Cd(OH)2).

Step 2: Calculate the moles of \(\mathrm{Cd^{2+}}\) ions in the solution

To find the moles of Cd^2+ ions, we can use the given volume (35.0 mL) and concentration (0.500 M) of the cadmium nitrate solution. Moles of Cd^2+ = Volume (L) × Concentration (M) First, convert the volume from milliliters to liters: \( 35.0 \mathrm{~mL} = 35.0 \times 10^{-3} \mathrm{~L} = 0.0350 \mathrm{~L} \) Then, calculate the moles of Cd^2+: Moles of Cd^2+ = \( 0.0350 \mathrm{~L} \times 0.500 \mathrm{~M} \) = 0.0175 moles

Step 3: Calculate moles of \(\mathrm{NaOH}\) needed

According to the balanced equation, 2 moles of NaOH are required to precipitate 1 mole of Cd^2+. Therefore, we can calculate the moles of NaOH needed: Moles of NaOH = \( 0.0175 \) moles of Cd^2+ × \( \dfrac{2 \ \mathrm{moles \ of \ NaOH}}{1 \ \mathrm{mole \ of \ Cd^{2+}}} \) = 0.0350 moles of NaOH

Step 4: Calculate mass of \(\mathrm{NaOH}\) needed

To find the mass of NaOH required, we'll use the molar mass of NaOH. The molar mass of NaOH can be calculated as follows: Molar mass of NaOH = \( (1 \times 22.99 \ \mathrm{g/mol})+(1 \times 15.999 \ \mathrm{g/mol})+(1 \times 1.0079 \ \mathrm{g/mol}) = 39.9969 \ \mathrm{g/mol}\) Now, we can calculate the mass of NaOH needed: Mass of NaOH = moles of NaOH × molar mass of NaOH Mass of NaOH = \( 0.0350 \mathrm{~moles} \times 39.9969 \mathrm{~g/mol} \) = 1.3998 g So, the mass of sodium hydroxide needed to precipitate the Cd^2+ ions from 35.0 mL of 0.500 M cadmium nitrate solution is approximately 1.40 grams.

Key Concepts

Understanding Chemical Equations

Chemical equations are essential in chemistry. They represent reactions that occur between substances. Writing a balanced chemical equation is crucial because it tells us the exact proportions of each substance involved in the reaction. In our problem, the equation:

• \( \mathrm{Cd(NO_3)_2 + 2NaOH \rightarrow Cd(OH)_2 + 2NaNO_3} \)

Made it clear that 2 moles of \( \mathrm{NaOH} \) are needed for every mole of \( \mathrm{Cd}^{2+} \) ions. This information is vital for determining how much \( \mathrm{NaOH} \) you will need to react completely with the \( \mathrm{Cd(NO_3)_2} \) solution. Always ensure that your chemical equations are balanced to accurately reflect the law of conservation of mass.

Calculating Molar Mass

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). Calculating the molar mass helps us convert between the mass of a substance and the amount in moles. To find the molar mass of a compound like \( \mathrm{NaOH} \), add up the molar masses of its elements:

• Sodium (Na): 22.99 g/mol
• Oxygen (O): 15.999 g/mol
• Hydrogen (H): 1.0079 g/mol

The total molar mass of \( \mathrm{NaOH} \) is 39.9969 g/mol. Knowing this lets us calculate the mass needed once we determine the number of moles required in a reaction.

Exploring Precipitation Reactions

In chemistry, a precipitation reaction occurs when two solutions combine to form an insoluble solid, known as a precipitate. These reactions typically happen when ions in aqueous solutions form a compound that is insoluble in water. In our scenario, mixing \( \mathrm{Cd(NO_3)_2} \) with \( \mathrm{NaOH} \) results in cadmium hydroxide \( \mathrm{Cd(OH)_2} \), the precipitate. Precipitation reactions are useful for removing or recovering substances from solutions. They are easy to recognize because the product is often a solid that settles out of the solution. Observing such reactions can provide evidence of chemical changes occurring during experiments.