H3PO4 + Fe(OH)3 = H2O + FePO4

H_3PO_4 phosphoric acid + Fe(OH)_3 iron(III) hydroxide ⟶ H_2O water + FePO_4 iron(III) phosphate

Balance the chemical equation algebraically: H_3PO_4 + Fe(OH)_3 ⟶ H_2O + FePO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 Fe(OH)_3 ⟶ c_3 H_2O + c_4 FePO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Fe: H: | 3 c_1 + 3 c_2 = 2 c_3 O: | 4 c_1 + 3 c_2 = c_3 + 4 c_4 P: | c_1 = c_4 Fe: | 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 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_3PO_4 + Fe(OH)_3 ⟶ 3 H_2O + FePO_4

+ ⟶ +

phosphoric acid + iron(III) hydroxide ⟶ water + iron(III) phosphate

Construct the equilibrium constant, K, expression for: H_3PO_4 + Fe(OH)_3 ⟶ H_2O + FePO_4 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: H_3PO_4 + Fe(OH)_3 ⟶ 3 H_2O + FePO_4 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 H_3PO_4 | 1 | -1 Fe(OH)_3 | 1 | -1 H_2O | 3 | 3 FePO_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_3PO_4 | 1 | -1 | ([H3PO4])^(-1) Fe(OH)_3 | 1 | -1 | ([Fe(OH)3])^(-1) H_2O | 3 | 3 | ([H2O])^3 FePO_4 | 1 | 1 | [FePO4] 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 = ([H3PO4])^(-1) ([Fe(OH)3])^(-1) ([H2O])^3 [FePO4] = (([H2O])^3 [FePO4])/([H3PO4] [Fe(OH)3])

Construct the rate of reaction expression for: H_3PO_4 + Fe(OH)_3 ⟶ H_2O + FePO_4 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: H_3PO_4 + Fe(OH)_3 ⟶ 3 H_2O + FePO_4 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 H_3PO_4 | 1 | -1 Fe(OH)_3 | 1 | -1 H_2O | 3 | 3 FePO_4 | 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 H_3PO_4 | 1 | -1 | -(Δ[H3PO4])/(Δt) Fe(OH)_3 | 1 | -1 | -(Δ[Fe(OH)3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) FePO_4 | 1 | 1 | (Δ[FePO4])/(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[H3PO4])/(Δt) = -(Δ[Fe(OH)3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[FePO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| phosphoric acid | iron(III) hydroxide | water | iron(III) phosphate formula | H_3PO_4 | Fe(OH)_3 | H_2O | FePO_4 Hill formula | H_3O_4P | FeH_3O_3 | H_2O | FeO_4P name | phosphoric acid | iron(III) hydroxide | water | iron(III) phosphate IUPAC name | phosphoric acid | ferric trihydroxide | water | iron(+3) cation phosphate