KOH + H3PO4 = H2O + K3PO4

KOH (potassium hydroxide) + H_3PO_4 (phosphoric acid) ⟶ H_2O (water) + K3PO4

Balance the chemical equation algebraically: KOH + H_3PO_4 ⟶ H_2O + K3PO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 H_3PO_4 ⟶ c_3 H_2O + c_4 K3PO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and P: H: | c_1 + 3 c_2 = 2 c_3 K: | c_1 = 3 c_4 O: | c_1 + 4 c_2 = c_3 + 4 c_4 P: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 KOH + H_3PO_4 ⟶ 3 H_2O + K3PO4

+ ⟶ + K3PO4

potassium hydroxide + phosphoric acid ⟶ water + K3PO4

Construct the equilibrium constant, K, expression for: KOH + H_3PO_4 ⟶ H_2O + K3PO4 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 KOH + H_3PO_4 ⟶ 3 H_2O + K3PO4 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 | 3 | -3 H_3PO_4 | 1 | -1 H_2O | 3 | 3 K3PO4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 3 | -3 | ([KOH])^(-3) H_3PO_4 | 1 | -1 | ([H3PO4])^(-1) H_2O | 3 | 3 | ([H2O])^3 K3PO4 | 1 | 1 | [K3PO4] 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])^(-3) ([H3PO4])^(-1) ([H2O])^3 [K3PO4] = (([H2O])^3 [K3PO4])/(([KOH])^3 [H3PO4])

Construct the rate of reaction expression for: KOH + H_3PO_4 ⟶ H_2O + K3PO4 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 KOH + H_3PO_4 ⟶ 3 H_2O + K3PO4 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 | 3 | -3 H_3PO_4 | 1 | -1 H_2O | 3 | 3 K3PO4 | 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 | 3 | -3 | -1/3 (Δ[KOH])/(Δt) H_3PO_4 | 1 | -1 | -(Δ[H3PO4])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) K3PO4 | 1 | 1 | (Δ[K3PO4])/(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[KOH])/(Δt) = -(Δ[H3PO4])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[K3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | phosphoric acid | water | K3PO4 formula | KOH | H_3PO_4 | H_2O | K3PO4 Hill formula | HKO | H_3O_4P | H_2O | K3O4P name | potassium hydroxide | phosphoric acid | water |

| potassium hydroxide | phosphoric acid | water | K3PO4 molar mass | 56.105 g/mol | 97.994 g/mol | 18.015 g/mol | 212.26 g/mol phase | solid (at STP) | liquid (at STP) | liquid (at STP) | melting point | 406 °C | 42.4 °C | 0 °C | boiling point | 1327 °C | 158 °C | 99.9839 °C | density | 2.044 g/cm^3 | 1.685 g/cm^3 | 1 g/cm^3 | solubility in water | soluble | very 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 | odorless |