Problem 110 - Chemistry The Central Science In Si Units

Question

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?

Step-by-Step Solution

Answer

(a) NH4+: N with 4 H; Cl-: Cl with 8 dots. (b) No N-Cl bond in solid NH4Cl. (c) 0.524 M. (d) 44.50 g AgNO3.

Step 1: Draw Lewis Structure for NH4+

The ammonium ion, NH4+, has a nitrogen atom bonded to four hydrogen atoms. Nitrogen has 5 valence electrons, and each hydrogen provides 1 electron, totaling 9 electrons. Since it's a cation with a +1 charge, we remove 1 electron. Distribute 8 electrons around nitrogen, forming single bonds with four hydrogen atoms, and give nitrogen a positive charge.

Step 2: Draw Lewis Structure for Cl-

The chloride ion, Cl-, has 7 valence electrons initially, and gains 1 electron to become Cl- with a full octet of 8 electrons. It thus has a -1 charge, and the Lewis structure is represented by a Cl atom surrounded by 8 dots, representing its valence electrons.

Step 3: Evaluate N-Cl Bond in Solid Ammonium Chloride

Ammonium chloride is composed of NH4+ and Cl- ions. In its solid state, these ions form an ionic lattice, rather than covalent bonds between N and Cl. Therefore, there is no direct N-Cl bond in solid ammonium chloride.

Step 4: Calculate Molar Mass of NH4Cl

Calculate the molar mass (MM) of NH4Cl: N (14.01 g/mol) + 4H (4 x 1.01 g/mol) + Cl (35.45 g/mol) = 53.49 g/mol.

Step 5: Calculate Concentration of NH4Cl Solution

To find the molar concentration, first convert the mass of NH4Cl to moles: \( n = \frac{14 \text{ g}}{53.49 \text{ g/mol}} \approx 0.262 \text{ mol} \). Then find the concentration by dividing moles by volume: \( C = \frac{0.262 \text{ mol}}{0.500 \text{ L}} = 0.524 \text{ M} \).

Step 6: Determine Mass of AgNO3 Needed

Use stoichiometry where \( 1 \text{ mol of } \text{Cl}^- \text{ reacts with } 1 \text{ mol of } \text{Ag}^+ \) to form AgCl. With a chloride concentration of 0.524 M in 0.500 L solution, moles of chloride = 0.262 mol. Calculate AgNO3 mass needed: \( 0.262 \text{ mol} \times 169.87 \text{ g/mol (molar mass of } \text{AgNO}_3 \text{)} = 44.50 \text{ g}.\)

Key Concepts

Lewis Structures

A Lewis structure is a diagrammatic way to represent the bonds between atoms and the valence electrons in a molecule. It plays an essential role in visualizing molecular structures. For the ammonium ion, denoted as \( \text{NH}_4^+ \), the nitrogen atom is centrally positioned, surrounded by four hydrogen atoms. Nitrogen starts with five valence electrons. Each hydrogen atom contributes one electron, leading to a total of nine electrons initially. However, due to the positive charge on the ammonium ion, we need to remove one electron, leaving eight electrons. These eight electrons form four single covalent bonds between the nitrogen and each hydrogen atom. This configuration results in nitrogen having a positive charge.

In the case of the chloride ion, \( \text{Cl}^- \), there are seven valence electrons around a chlorine atom initially. By gaining one electron, chloride achieves a stable octet, making it a negatively charged ion. The Lewis structure of \( \text{Cl}^- \) displays a chlorine atom surrounded by eight dots, representing its full octet of valence electrons.

Ionic Compounds

Ionic compounds are formed when positively charged ions bond with negatively charged ions. Ammonium chloride, \( \text{NH}_4\text{Cl} \), is an interesting example as it combines a polyatomic ion, \( \text{NH}_4^+ \), with a simple anion, \( \text{Cl}^- \). In ionic compounds such as ammonium chloride, each ion retains its identity, forming a lattice structure in solid form rather than sharing electron-pairs as seen in covalent bonds.

• In solid ammonium chloride, there's no direct molecule-to-molecule bond between nitrogen in ammonium and chlorine in chloride.
• These ions are held together by the strong electrostatic forces of attraction between oppositely charged ions.
• This makes ammonium chloride dissolve well in water, separating into \( \text{NH}_4^+ \) and \( \text{Cl}^- \) ions.

Understanding the nature of these ionic interactions is crucial for predicting the behavior of compounds in different conditions.

Molar Concentration

Molar concentration, or molarity, is a measure of the concentration of a solute in a solution. It is expressed in moles of solute per liter of solution. For ammonium chloride, if you dissolve 14 grams of this salt in 500.0 mL of water, you can determine the molar concentration by following these steps:

• First, identify its molar mass, which is 53.49 g/mol. Add together the molar masses: nitrogen (14.01 g/mol), hydrogen (1.01 g/mol per hydrogen times four), and chlorine (35.45 g/mol).
• Convert the mass of ammonium chloride to moles: \( n = \frac{14 \text{ g}}{53.49 \text{ g/mol}} \approx 0.262 \text{ mol} \).
• The volume of the solution is 0.500 L. Divide the moles by the volume to find molarity: \( C = \frac{0.262 \text{ mol}}{0.500 \text{ L}} = 0.524 \text{ M} \).

This provides a clear understanding of how much solute is present in a given volume of solution.

Precipitation Reaction

Precipitation reactions result in the formation of an insoluble solid. They occur when two aqueous solutions react and one of the new compounds formed is not soluble in water. In this exercise, silver nitrate \( \text{AgNO}_3 \) reacts with ammonium chloride solution to form silver chloride \( \text{AgCl} \) which precipitates out as a solid.

The reaction follows this stoichiometry: \[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} (s) \]
**To carry out the reaction:**

• With a chloride ion concentration of 0.524 M in 0.500 L of solution, calculate moles of chloride: 0.262 mol.
• The reaction with \( \text{AgNO}_3 \) is in a 1:1 ratio.
• Calculate the required mass of \( \text{AgNO}_3 \) to precipitate all chloride ions as silver chloride: \( 0.262 \text{ mol} \times 169.87 \text{ g/mol (molar mass of } \text{AgNO}_3 \text{)} = 44.50 \text{ g} \).

Precipitation reactions such as this are valuable in driving chemical reactions to completion, removing ions from solution.