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

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, \(1.40 \mathrm{g}\) of \(\mathrm{NaOH}\) is needed.

Step-by-step solution

1. Write the balanced chemical equation for the reaction between NaOH and Cd(NO3)₂

The balanced chemical equation for the reaction between sodium hydroxide \(\left(\mathrm{NaOH}\right)\) and cadmium nitrate \(\left[\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\right]\) is given by: \[ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\left(aq\right) + 2\mathrm{NaOH}\left(aq\right) \rightarrow \mathrm{Cd(OH)}_{2}\left(s\right) + 2\mathrm{NaNO}_{3}\left(aq\right) \]

2. Determine the moles of Cd(NO3)₂ in the solution

Given the volume and the concentration of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) solution, we can calculate the number of moles present using the formula: moles \(= \) volume \( \times \) concentration moles of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} = 35.0 \mathrm{~mL} \times 0.500 \mathrm{M}\) Since \(\mathrm{1~L = 1000~mL}\): moles of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} = \dfrac{35.0}{1000} \mathrm{L} \times 0.500 \mathrm{M}\) moles of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} = 0.0175 \mathrm{~mol}\)

3. Determine the moles of NaOH needed for the reaction

From the balanced chemical equation, we see that 1 mole of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) reacts with 2 moles of \(\mathrm{NaOH}\). Hence, we can determine the moles of \(\mathrm{NaOH}\) by using the stoichiometry of the reaction: moles of \(\mathrm{NaOH} = \) moles of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) \(\times\) \(\dfrac{2 \mathrm{~moles\ NaOH}}{1\ \mathrm{mole\ Cd(NO}_{3})_{2}}\) moles of \(\mathrm{NaOH} = 0.0175 \mathrm{~mol} \times \dfrac{2}{1}\) moles of \(\mathrm{NaOH} = 0.0350 \mathrm{~mol}\)

4. Calculate the mass of NaOH needed

Now that we know the number of moles of \(\mathrm{NaOH}\) required, we can calculate the mass using the molar mass of \(\mathrm{NaOH}\): mass \(= \) moles \( \times \) molar mass The molar mass of \(\mathrm{NaOH}\) is \(22.99\ \mathrm{g/mol\ Na} + 15.99\ \mathrm{g/mol\ O} + 1.008\ \mathrm{g/mol\ H} = 40.00\ \mathrm{g/mol}\) Hence, the mass of \(\mathrm{NaOH}\) needed is: mass of \(\mathrm{NaOH} = 0.0350 \mathrm{~mol} \times 40.00 \mathrm{g/mol}\) mass of \(\mathrm{NaOH} = 1.40 \mathrm{g}\) So, \(1.40\ \mathrm{g}\) 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.