Al(OH)3 + HF = H2O + AlF3

Al(OH)_3 aluminum hydroxide + HF hydrogen fluoride ⟶ H_2O water + AlF_3 aluminum fluoride

Balance the chemical equation algebraically: Al(OH)_3 + HF ⟶ H_2O + AlF_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al(OH)_3 + c_2 HF ⟶ c_3 H_2O + c_4 AlF_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Al, H, O and F: Al: | c_1 = c_4 H: | 3 c_1 + c_2 = 2 c_3 O: | 3 c_1 = c_3 F: | c_2 = 3 c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Al(OH)_3 + 3 HF ⟶ 3 H_2O + AlF_3

+ ⟶ +

aluminum hydroxide + hydrogen fluoride ⟶ water + aluminum fluoride

Construct the equilibrium constant, K, expression for: Al(OH)_3 + HF ⟶ H_2O + AlF_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: Al(OH)_3 + 3 HF ⟶ 3 H_2O + AlF_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i Al(OH)_3 | 1 | -1 HF | 3 | -3 H_2O | 3 | 3 AlF_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al(OH)_3 | 1 | -1 | ([Al(OH)3])^(-1) HF | 3 | -3 | ([HF])^(-3) H_2O | 3 | 3 | ([H2O])^3 AlF_3 | 1 | 1 | [AlF3] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([Al(OH)3])^(-1) ([HF])^(-3) ([H2O])^3 [AlF3] = (([H2O])^3 [AlF3])/([Al(OH)3] ([HF])^3)

Construct the rate of reaction expression for: Al(OH)_3 + HF ⟶ H_2O + AlF_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: Al(OH)_3 + 3 HF ⟶ 3 H_2O + AlF_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i Al(OH)_3 | 1 | -1 HF | 3 | -3 H_2O | 3 | 3 AlF_3 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term Al(OH)_3 | 1 | -1 | -(Δ[Al(OH)3])/(Δt) HF | 3 | -3 | -1/3 (Δ[HF])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) AlF_3 | 1 | 1 | (Δ[AlF3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -(Δ[Al(OH)3])/(Δt) = -1/3 (Δ[HF])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[AlF3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| aluminum hydroxide | hydrogen fluoride | water | aluminum fluoride formula | Al(OH)_3 | HF | H_2O | AlF_3 Hill formula | AlH_3O_3 | FH | H_2O | AlF_3 name | aluminum hydroxide | hydrogen fluoride | water | aluminum fluoride IUPAC name | aluminum hydroxide | hydrogen fluoride | water | trifluoroalumane

| aluminum hydroxide | hydrogen fluoride | water | aluminum fluoride molar mass | 78.003 g/mol | 20.006 g/mol | 18.015 g/mol | 83.976748 g/mol phase | | gas (at STP) | liquid (at STP) | solid (at STP) melting point | | -83.36 °C | 0 °C | 1040 °C boiling point | | 19.5 °C | 99.9839 °C | density | | 8.18×10^-4 g/cm^3 (at 25 °C) | 1 g/cm^3 | 3.1 g/cm^3 solubility in water | | miscible | | surface tension | | | 0.0728 N/m | dynamic viscosity | | 1.2571×10^-5 Pa s (at 20 °C) | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |