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.
