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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
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

01

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).
02

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
03

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
04

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.

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Most popular questions from this chapter

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) \(\mathrm{H}_{2} \mathrm{SO}_{3}\), (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethanol), (c) \(\mathrm{NH}_{3}\), (d) \(\mathrm{KClO}_{3}\), (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\)

Using modern analytical techniques, it is possible to detect sodium ions in concentrations as low as \(50 \mathrm{pg} / \mathrm{mL}\). What is this detection limit expressed in (a) molarity of \(\mathrm{Na}^{+}\) (b) \(\mathrm{Na}^{+}\) ions per cubic centimeter?

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

When methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) is dissolved in water, a nonconducting solution results. When acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) dissolves in water, the solution is weakly conducting and acidic in nature. Describe what happens upon dissolution in the two cases, and account for the different results.
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