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?

Short 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-by-step solution

Write the balanced chemical equation

The balanced equation for the reaction between sodium hydroxide and cadmium nitrate is: $\mathrm{Cd}({\mathrm{NO}}_3{)}_2+2\mathrm{NaOH}\to \mathrm{Cd}(\mathrm{OH}{)}_2+2{\mathrm{NaNO}}_3$ 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).

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

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+ × ${\displaystyle \frac{2\mathrm{moles}\mathrm{of}\mathrm{NaOH}}{1\mathrm{mole}\mathrm{of}{\mathrm{Cd}}^{2+}}}$ = 0.0350 moles of NaOH

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 = $(1\times 22.99\mathrm{g}/\mathrm{mol})+(1\times 15.999\mathrm{g}/\mathrm{mol})+(1\times 1.0079\mathrm{g}/\mathrm{mol})=39.9969\mathrm{g}/\mathrm{mol}$ Now, we can calculate the mass of NaOH needed: Mass of NaOH = moles of NaOH × molar mass of NaOH Mass of NaOH = $0.0350\mathrm{moles}\times 39.9969\mathrm{g}/\mathrm{mol}$ = 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:

• $\mathrm{Cd}({\mathrm{NO}}_3{)}_2+2\mathrm{NaOH}\to \mathrm{Cd}(\mathrm{OH}{)}_2+2{\mathrm{NaNO}}_3$

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 $\mathrm{Cd}({\mathrm{NO}}_3{)}_2$ 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 $\mathrm{Cd}({\mathrm{NO}}_3{)}_2$ with $\mathrm{NaOH}$ results in cadmium hydroxide $\mathrm{Cd}(\mathrm{OH}{)}_2$, 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.