KOH + Al(OH)3 = H2O + KAlO2

KOH potassium hydroxide + Al(OH)_3 aluminum hydroxide ⟶ H_2O water + KAlO2

Balance the chemical equation algebraically: KOH + Al(OH)_3 ⟶ H_2O + KAlO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Al(OH)_3 ⟶ c_3 H_2O + c_4 KAlO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and Al: H: | c_1 + 3 c_2 = 2 c_3 K: | c_1 = c_4 O: | c_1 + 3 c_2 = c_3 + 2 c_4 Al: | c_2 = 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 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | KOH + Al(OH)_3 ⟶ 2 H_2O + KAlO2

+ ⟶ + KAlO2

potassium hydroxide + aluminum hydroxide ⟶ water + KAlO2

Construct the equilibrium constant, K, expression for: KOH + Al(OH)_3 ⟶ H_2O + KAlO2 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: KOH + Al(OH)_3 ⟶ 2 H_2O + KAlO2 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 KOH | 1 | -1 Al(OH)_3 | 1 | -1 H_2O | 2 | 2 KAlO2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 1 | -1 | ([KOH])^(-1) Al(OH)_3 | 1 | -1 | ([Al(OH)3])^(-1) H_2O | 2 | 2 | ([H2O])^2 KAlO2 | 1 | 1 | [KAlO2] 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 = ([KOH])^(-1) ([Al(OH)3])^(-1) ([H2O])^2 [KAlO2] = (([H2O])^2 [KAlO2])/([KOH] [Al(OH)3])

Construct the rate of reaction expression for: KOH + Al(OH)_3 ⟶ H_2O + KAlO2 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: KOH + Al(OH)_3 ⟶ 2 H_2O + KAlO2 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 KOH | 1 | -1 Al(OH)_3 | 1 | -1 H_2O | 2 | 2 KAlO2 | 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 KOH | 1 | -1 | -(Δ[KOH])/(Δt) Al(OH)_3 | 1 | -1 | -(Δ[Al(OH)3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) KAlO2 | 1 | 1 | (Δ[KAlO2])/(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[KOH])/(Δt) = -(Δ[Al(OH)3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[KAlO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | aluminum hydroxide | water | KAlO2 formula | KOH | Al(OH)_3 | H_2O | KAlO2 Hill formula | HKO | AlH_3O_3 | H_2O | AlKO2 name | potassium hydroxide | aluminum hydroxide | water |

| potassium hydroxide | aluminum hydroxide | water | KAlO2 molar mass | 56.105 g/mol | 78.003 g/mol | 18.015 g/mol | 98.078 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 406 °C | | 0 °C | boiling point | 1327 °C | | 99.9839 °C | density | 2.044 g/cm^3 | | 1 g/cm^3 | solubility in water | soluble | | | surface tension | | | 0.0728 N/m | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |