Ca(OH)2 + H3IO4 = H2O + Ca3(IO4)2

Input interpretation

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Ca(OH)_2 calcium hydroxide + H3IO4 ⟶ H_2O water + Ca3(IO4)2

Balanced equation

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Balance the chemical equation algebraically: Ca(OH)_2 + H3IO4 ⟶ H_2O + Ca3(IO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Ca(OH)_2 + c_2 H3IO4 ⟶ c_3 H_2O + c_4 Ca3(IO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Ca, H, O and I: Ca: | c_1 = 3 c_4 H: | 2 c_1 + 3 c_2 = 2 c_3 O: | 2 c_1 + 4 c_2 = c_3 + 8 c_4 I: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 2 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Ca(OH)_2 + 2 H3IO4 ⟶ 6 H_2O + Ca3(IO4)2

+ H3IO4 ⟶ + Ca3(IO4)2

Names

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calcium hydroxide + H3IO4 ⟶ water + Ca3(IO4)2

Equilibrium constant

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Construct the equilibrium constant, K, expression for: Ca(OH)_2 + H3IO4 ⟶ H_2O + Ca3(IO4)2 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: 3 Ca(OH)_2 + 2 H3IO4 ⟶ 6 H_2O + Ca3(IO4)2 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 Ca(OH)_2 | 3 | -3 H3IO4 | 2 | -2 H_2O | 6 | 6 Ca3(IO4)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Ca(OH)_2 | 3 | -3 | ([Ca(OH)2])^(-3) H3IO4 | 2 | -2 | ([H3IO4])^(-2) H_2O | 6 | 6 | ([H2O])^6 Ca3(IO4)2 | 1 | 1 | [Ca3(IO4)2] 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 = ([Ca(OH)2])^(-3) ([H3IO4])^(-2) ([H2O])^6 [Ca3(IO4)2] = (([H2O])^6 [Ca3(IO4)2])/(([Ca(OH)2])^3 ([H3IO4])^2)

Rate of reaction

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Construct the rate of reaction expression for: Ca(OH)_2 + H3IO4 ⟶ H_2O + Ca3(IO4)2 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: 3 Ca(OH)_2 + 2 H3IO4 ⟶ 6 H_2O + Ca3(IO4)2 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 Ca(OH)_2 | 3 | -3 H3IO4 | 2 | -2 H_2O | 6 | 6 Ca3(IO4)2 | 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 Ca(OH)_2 | 3 | -3 | -1/3 (Δ[Ca(OH)2])/(Δt) H3IO4 | 2 | -2 | -1/2 (Δ[H3IO4])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Ca3(IO4)2 | 1 | 1 | (Δ[Ca3(IO4)2])/(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 = -1/3 (Δ[Ca(OH)2])/(Δt) = -1/2 (Δ[H3IO4])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[Ca3(IO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)