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?

For the ammonium ion (NH₄⁺), 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⁻), chlorine (Cl) has 7 valence electrons. It gains 1 electron to complete its octet and becomes negatively charged: \[\mathrm{Cl}^{-} : \hspace{2mm} \ominus \]

In solid ammonium chloride (NH₄Cl), there is no direct N-Cl bond. Instead, the compound exists as an ionic lattice structure where the positively charged ammonium ions (NH₄⁺) are attracted to the negatively charged chloride ions (Cl⁻).

To calculate the molar concentration of the solution, first find the moles of ammonium chloride dissolved. The molar mass of NH₄Cl 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 ≈ 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 ≈ 0.524 M

When silver nitrate (AgNO₃) reacts with ammonium chloride (NH₄Cl), it forms silver chloride (AgCl) and ammonium nitrate (NH₄NO₃): \[ \mathrm{AgNO}_{3} (aq) + \mathrm{NH}_{4}\mathrm{Cl} (aq) \to \mathrm{AgCl}(s) + \mathrm{NH}_{4}\mathrm{NO}_{3} (aq) \] Notice that the reaction has a one-to-one stoichiometry for AgNO₃ and NH₄Cl. Since we know the moles of NH₄Cl in the solution (0.262 mol), we can find the moles of AgNO₃ needed: Moles of AgNO₃ = 0.262 mol The molar mass of AgNO₃ is 169.87 g/mol (107.87 for Ag, 14.01 for N, and 48.00 for O₃). Therefore, the mass of AgNO₃ needed: Mass of AgNO₃ = moles x molar mass = 0.262 mol x 169.87 g/mol ≈ 44.5 g So, 44.5 grams of silver nitrate is needed to precipitate all chloride as silver chloride.