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Ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a very soluble salt in water. (a) Draw the Lewis structures of the ammonium and chloride ions. (b) Is there an \(\mathrm{N}-\mathrm{Cl}\) bond in solid ammonium chloride? (c) If you dissolve \(14 \mathrm{~g}\) of ammonium chloride in \(500.0 \mathrm{~mL}\) of water, what is the molar concentration of the solution? (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?

Short Answer

Expert verified
(a) Ammonium ion: \(\mathrm{H} - \mathrm{N}^{+} - \mathrm{H}\) | \(\mathrm{H}\) and Chloride ion: \(\mathrm{Cl}^{-} : \hspace{2mm} \ominus \) (b) No, there is no direct N-Cl bond in solid ammonium chloride. (c) The molar concentration is approximately 0.524 M. (d) 44.5 grams of silver nitrate are needed to precipitate all chloride as silver chloride.

Step by step solution

01

Drawing Lewis structures of ammonium and chloride ions

For the ammonium ion (NH鈧勨伜), nitrogen (N) has 5 valence electrons and each hydrogen (H) has 1 valence electron. In the ammonium ion, nitrogen forms four covalent bonds with hydrogen atoms and has a positive charge: \[ \mathrm{H} - \mathrm{N}^{+} - \mathrm{H} \] | \(\mathrm{H}\) For the chloride ion (Cl鈦), chlorine (Cl) has 7 valence electrons. It gains 1 electron to complete its octet and becomes negatively charged: \[\mathrm{Cl}^{-} : \hspace{2mm} \ominus \]
02

N-Cl bond in solid ammonium chloride

In solid ammonium chloride (NH鈧凜l), there is no direct N-Cl bond. Instead, the compound exists as an ionic lattice structure where the positively charged ammonium ions (NH鈧勨伜) are attracted to the negatively charged chloride ions (Cl鈦).
03

Calculating the molar concentration of the solution

To calculate the molar concentration of the solution, first find the moles of ammonium chloride dissolved. The molar mass of NH鈧凜l is 53.49 g/mol (14.01 for N, 4 x 1.01 for each H, and 35.45 for Cl): Moles of ammonium chloride = mass / molar mass = 14 g / 53.49 g/mol 鈮 0.262 mol The volume of the solution is 500.0 mL, which is equal to 0.500 L. To calculate the molar concentration, divide the moles of solute by the volume of the solution: Molar concentration = 0.262 mol / 0.500 L 鈮 0.524 M
04

Determining the mass of silver nitrate needed to precipitate all chloride

When silver nitrate (AgNO鈧) reacts with ammonium chloride (NH鈧凜l), it forms silver chloride (AgCl) and ammonium nitrate (NH鈧凬O鈧): \[ \mathrm{AgNO}_{3} (aq) + \mathrm{NH}_{4}\mathrm{Cl} (aq) \to \mathrm{AgCl}(s) + \mathrm{NH}_{4}\mathrm{NO}_{3} (aq) \] Notice that the reaction has a one-to-one stoichiometry for AgNO鈧 and NH鈧凜l. Since we know the moles of NH鈧凜l in the solution (0.262 mol), we can find the moles of AgNO鈧 needed: Moles of AgNO鈧 = 0.262 mol The molar mass of AgNO鈧 is 169.87 g/mol (107.87 for Ag, 14.01 for N, and 48.00 for O鈧). Therefore, the mass of AgNO鈧 needed: Mass of AgNO鈧 = moles x molar mass = 0.262 mol x 169.87 g/mol 鈮 44.5 g So, 44.5 grams of silver nitrate is needed to precipitate all chloride as silver chloride.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structures
Lewis structures are diagrams that show the bonding between atoms of a molecule. This visual representation helps in understanding how electrons are shared between atoms. For the ammonium ion, NH鈧勨伜, nitrogen begins with five valence electrons. Each hydrogen contributes one electron for bonding. Thus, nitrogen forms four single covalent bonds with four hydrogen atoms. The NH鈧 molecule carries a positive charge because it has one less electron than the combined total of electrons from nitrogen and the hydrogens. In a Lewis structure, this can be depicted as:
  • The central nitrogen atom bonded to four hydrogen atoms in a tetrahedral shape.
  • The charge is indicated by a superscript plus sign (+) outside the brackets of the structure.

For the chloride ion, Cl鈦, chlorine has seven valence electrons. By gaining an electron, it achieves a stable octet. Thus, Cl becomes a chloride ion with a negative charge, shown as Cl鈦 with an extra electron represented alongside the symbol. The overall complete octet for chlorine is achieved with eight electrons around it: seven original from chlorine and the additional one gained electron.
Molar Concentration
Molar concentration (also called molarity) is a measure of the concentration of a solute in a solution. It is expressed in moles per liter ( ext{mol/L}"). To calculate it, first determine the number of moles of solute in the solution. For instance, if there are 14 grams of ammonium chloride (NH鈧凜l) dissolved, and the molar mass of NH鈧凜l is 53.49 g/mol, then:
  • Number of moles = mass / molar mass = 14 g / 53.49 g/mol 鈮 0.262 mol

Then, you measure the volume of the solution in liters. For 500.0 mL of water, convert it to liters: 500.0 mL is equivalent to 0.500 L.
  • Molarity (M) = moles of solute / volume of solution in liters = 0.262 mol / 0.500 L 鈮 0.524 M

This calculation indicates that the solution of ammonium chloride has a concentration of 0.524 mol/L.
Ionic Lattice Structure
An ionic lattice structure is a highly ordered, repeating pattern in which ions are arranged. In a solid ammonium chloride sample, the NH鈧勨伜 and Cl鈦 ions do not form a direct nitrogen-chlorine bond. Instead, they align to create a crystal lattice. Here, each positively charged ammonium ion is surrounded by negatively charged chloride ions. The key characteristics of an ionic lattice structure include:
  • The arrangement is maintained by strong electrostatic forces of attraction between oppositely charged ions.
  • The structure is usually very stable and solid at room temperature.
  • It leads to distinct properties such as high melting and boiling points.

This orderly structure explains why solid ammonium chloride has no specific N-Cl bond, but rather exists as an ionic compound.
Precipitation Reaction
A precipitation reaction involves mixing two solutions that result in the formation of an insoluble solid, known as the precipitate. In the context of ammonium chloride, if silver nitrate (AgNO鈧) is added to the NH鈧凜l solution, silver chloride (AgCl) precipitates out. The detailed reaction is:
  • AgNO鈧(aq) + NH鈧凜l(aq) 鈫 AgCl(s) + NH鈧凬O鈧(aq)

In this reaction:
  • Silver ions (Ag鈦) from silver nitrate react with chloride ions (Cl鈦) from ammonium chloride to form AgCl, which is insoluble in water.
  • AgCl appears as a whitish solid that settles out of the solution.
  • The remaining ions, NH鈧勨伜 and NO鈧冣伝, stay dissolved in water as ammonium nitrate.

The stoichiometry of the reaction is one-to-one: one mole of silver nitrate reacts with one mole of ammonium chloride to produce one mole of silver chloride, making it crucial to calculate the correct mass of silver nitrate needed for complete precipitation based on the initial moles of chloride ions present.

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Most popular questions from this chapter

State whether each of these statements is true or false. (a) The longer the bond, the stronger the bond. (b) \(\mathrm{C}-\mathrm{C}\) bonds are stronger than \(\mathrm{C}-\mathrm{F}\) bonds. (c) A typical double bond length is in the \(500-1000\) pm range. (d) Energy is required to form a chemical bond. (e) The longer the bond, the more energy is stored chemical bonds.

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7. (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. \((\mathbf{b})\) The \(\mathrm{B}-\mathrm{N}\) bond is obviously more polar than the C-C bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4 ). (c) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(\mathrm{B}-\mathrm{N}\) bond in the trichloride reaction product would be more or less polar than the \(\mathrm{B}-\mathrm{N}\) bond in the trifluoride product, and justify your reasoning.

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: \((\mathbf{a}) \mathrm{PF}_{6}^{-},(\mathbf{b}) \mathrm{BeCl}_{2},(\mathbf{c}) \mathrm{NH}_{3},(\mathbf{d}) \mathrm{XeF}_{2} \mathrm{O}\) (the Xe is the central atom), (e) \(\mathrm{SO}_{4}^{2-}\).

A common form of elemental phosphorus is the white phosphorus, where four \(\mathrm{P}\) atoms are arranged in a tetrahedron. All four phosphorus atoms are equivalent. White phosphorus reacts spontaneously with the oxygen in air to form \(\mathrm{P}_{4} \mathrm{O}_{6} .\) (a) How many valance electron pairs are in the \(\mathrm{P}_{4} \mathrm{O}_{6}\) molecule? (b) When \(\mathrm{P}_{4} \mathrm{O}_{6}\) is dissolved in water, it produces a \(\mathrm{H}_{3} \mathrm{PO}_{3}\), molecule. \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has two forms, \(\mathrm{P}\) forms 3 covalent bonds in the first form and \(\mathrm{P}\) forms 5 covalent bonds in the second form. Draw two possible Lewis structures of \(\mathrm{H}_{3} \mathrm{PO}_{3}\). (c) Which structure obeys the octet rule?

Potassium peroxide is composed of \(70.96 \% \mathrm{~K}\) and \(29.04 \% \mathrm{O}\). Each peroxide ion has a net charge of \(2^{-}\). (a) Given the molecular weight of the compound is 110.19 , determine the chemical formula of potassium peroxide. \((\mathbf{b})\) What is the number of valance electron pairs in the peroxide ion? (c) Draw the Lewis structure of the peroxide ion. (d) Compare the \(\mathrm{O}-\mathrm{O}\) bond distance in oxygen molecule and potassium peroxide.

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