You have two 500.0 ml aqueous solutions. solution a is a solution of a metal nitrate that is 8.246% nitrogen by mass the ionic compound in solution b consists of potassium, chromium, and oxygen; chromium has an oxidation state of + 6 and there are two potassiums and 1 chromium in the formula. the masses of the solutes in each solution are the same. when the solutions are added together, a blood-red precipitate forms. after the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g. identify the ionic compounds in solution a and solution

b. identify the blood-red precipitate. calculate the concentration (molarity) of all ions in the original solutions. calculate the concentration (molarity) of all ions in the final solution.

1) Answer is: the ionic compound in the solution b is K₂CrO₄ (potassium chromate).

Ionic compound in solution b has two potassiums (oxidation number + 1), one chromium (oxidation number + 6) and four oxygens. Oxidation number of oxygen is - 2 and compound has neutral charge:

2 · (+1) + 6 + x · (-2) = 0.

x = 4; number of oxygen atoms.

2) Answer is: the ionic compound in solution a is AgNO₃ (silver nitrate).

ω (N) = 8.246% : 100%.

ω (N) = 0.08246; mass percentage of nitrogen.

M (MNO₃) = M (N) : ω (N).

M (MNO₃) = 14 g/mol : 0.08246.

M (MNO₃) = 169.8 g/mol; molar mass of metal nitrate.

M (M) = M (MNO₃) - M (N) - 3 · M (O).

M (M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol.

M (M) = 107.8 g/mol; atomic mass of metal, this metal is silver (Ag).

3) Balanced chemical reaction:

2AgNO₃ (aq) + K₂CrO₄ (aq) → Ag₂CrO₄ (s) + 2KNO₃ (aq).

Ionic reaction:

2Ag⁺ (aq) + 2NO₃ (aq) + 2K⁺ (aq) + CrO₄²⁻ (aq) → Ag₂CrO₄ (s) + 2K⁺ (aq) + 2NO₃⁻ (aq).

Net ionic reaction: 2Ag⁺ (aq) + CrO₄²⁻ (aq) → Ag₂CrO₄ (s).

Answer is: the blood-red precipitate is silver chromate (Ag₂CrO₄).

4) m (Ag₂CrO₄) = 331.8 g; mass of solid silver chromate.

n (Ag₂CrO₄) = m (Ag₂CrO₄) : M (Ag₂CrO₄).

n (Ag₂CrO₄) = 331.8 g : 331.8 g/mol.

n (Ag₂CrO₄) = 1 mol; amount of silver chromate.

From balanced chemical reaction: n (Ag₂CrO₄) : n (AgNO₃) = 1 : 2.

n (AgNO₃) = 2 · 1 mol.

n (AgNO₃) = 2 mol.

m (AgNO₃) = n (AgNO₃) · M (AgNO₃).

m (AgNO₃) = 2 mol · 169.8 g/mol.

m (AgNO₃) = 339.6 g; mass of silver nitrate.

m (AgNO₃) = m (K₂CrO₄).

m (K₂CrO₄) = 339.6 g; mass of potassium chromate.

n (K₂CrO₄) = m (K₂CrO₄) : M (K₂CrO₄).

n (K₂CrO₄) = 339.6 g : 194.2 g/mol.

n (K₂CrO₄) = 1.75 mol; amount of potassium chromate.

5) Chemical reaction of dissociation of silver nitrate in water:

AgNO₃ (aq) → Ag⁺ (aq) + NO₃⁻ (aq).

V (solution a) = 500 mL : 1000 mL/L.

V (solution a) = 0.5 L; volume of solution a.

c (AgNO₃) = n (AgNO₃) : V (solution a).

c (AgNO₃) = 2 mol : 0.5 L.

c (AgNO₃) = 4 mol/L = 4 M.

From dissociation of silver nitrate: c (AgNO₃) = c (Ag⁺) = c (NO₃⁻).

c (Ag⁺) = 4 M; the concentration of silver ions in the original solution a.

c (NO₃⁻) = 4 M; the concentration of silver ions in the original solution a.

6) Chemical reaction of dissociation of potssium chromate in water:

K₂CrO₄ (aq) → 2K⁺ (aq) + CrO₄²⁻ (aq).

V (solution b) = 500 mL : 1000 mL/L.

V (solution b) = 0.5 L; volume of solution b.

c (K₂CrO₄) = n (K₂CrO₄) : V (solution b).

c (AgNO₃) = 1.75 mol : 0.5 L.

c (AgNO₃) = 3.5 mol/L = 3.5 M.

From dissociation of silver nitrate: c (K₂CrO₄) = c/2 (K⁺) = c (CrO₄²⁻).

c (K⁺) = 7 M; the concentration of potassium ions in the original solution b.

c (CrO₄²⁻) = 3.5 M; the concentration of silver ions in the original solution b.

7) V (final solution) = V (solution a) + V (solution b).

V (final solution) = 500.0 mL + 500.0 mL.

V (final solution) = 1000 mL : 1000 mL/L.

V (final solution) = 1 L.

n (NO₃⁻) = 2 mol.

c (NO₃⁻) = n (NO₃⁻) : V (final solution)

c (NO₃⁻) = 2 mol : 1 L.

c (NO₃⁻) = 2 M; the concentration of nitrate anions in final solution.

8) in the solution b there were 3.5 mol of potassium cations, but one part of them reacts with 2 moles of nitrate anions:

K⁺ (aq) + NO₃⁻ (aq) → KNO₃ (aq).

From chemical reaction: n (K⁺) : n (NO₃⁻) = 1 : 1.

Δn (K⁺) = 3.5 mol - 2 mol.

Δn (K⁺) = 1.5 mol; amount of potassium anions left in final solution.

c (K⁺) = Δn (K⁺) : V (final solution).

c (K⁺) = 1.5 mol : 1 L.

c (K⁺) = 1.5 M; the concentration of potassium cations in final solution.