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Chapter 4: Problem 80

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

Expert verified
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

01

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

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: \[ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\left(aq\right) + 2\mathrm{NaOH}\left(aq\right) \rightarrow \mathrm{Cd(OH)}_{2}\left(s\right) + 2\mathrm{NaNO}_{3}\left(aq\right) \]
02

2. Determine the moles of Cd(NO3)鈧� 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 \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} = 0.0175 \mathrm{~mol}\)
03

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 \(\mathrm{NaOH} = 0.0350 \mathrm{~mol}\)
04

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.

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Most popular questions from this chapter

Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg} \mathrm{SO}_{4}(a q)\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{Na}_{2} \mathrm{~S}(a q)\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\) and \(\mathrm{CaCl}_{2}(a q)\)

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

(a) Calculate the molarity of a solution made by dissolving \(0.750\) grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in enough water to form exactly \(850 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{KMnO}_{4}\) are present in \(250 \mathrm{~mL}\) of a \(0.0475 \mathrm{M}\) solution? (c) How many milliliters of \(11.6 \mathrm{M} \mathrm{HCl}\) solution are needed to obtain \(0.250 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

Suppose you have a solution that might contain any or. all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+}\), and \(\mathrm{Mn}^{2+}\). Addition of \(\mathrm{HCl}\) solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resultant solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

When asked what causes electrolyte solutions to conduct electricity, a student responds that it is due to the movement of electrons through the solution. Is the student correct? If not, what is the correct response?
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