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Ba(OH)2 + K3PO4 = KOH + Ba3(PO4)2

Input interpretation

Ba(OH)_2 barium hydroxide + K3PO4 ⟶ KOH potassium hydroxide + Ba3(PO4)2
Ba(OH)_2 barium hydroxide + K3PO4 ⟶ KOH potassium hydroxide + Ba3(PO4)2

Balanced equation

Balance the chemical equation algebraically: Ba(OH)_2 + K3PO4 ⟶ KOH + Ba3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Ba(OH)_2 + c_2 K3PO4 ⟶ c_3 KOH + c_4 Ba3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Ba, H, O, K and P: Ba: | c_1 = 3 c_4 H: | 2 c_1 = c_3 O: | 2 c_1 + 4 c_2 = c_3 + 8 c_4 K: | 3 c_2 = c_3 P: | 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 Ba(OH)_2 + 2 K3PO4 ⟶ 6 KOH + Ba3(PO4)2
Balance the chemical equation algebraically: Ba(OH)_2 + K3PO4 ⟶ KOH + Ba3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Ba(OH)_2 + c_2 K3PO4 ⟶ c_3 KOH + c_4 Ba3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Ba, H, O, K and P: Ba: | c_1 = 3 c_4 H: | 2 c_1 = c_3 O: | 2 c_1 + 4 c_2 = c_3 + 8 c_4 K: | 3 c_2 = c_3 P: | 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 Ba(OH)_2 + 2 K3PO4 ⟶ 6 KOH + Ba3(PO4)2

Structures

+ K3PO4 ⟶ + Ba3(PO4)2
+ K3PO4 ⟶ + Ba3(PO4)2

Names

barium hydroxide + K3PO4 ⟶ potassium hydroxide + Ba3(PO4)2
barium hydroxide + K3PO4 ⟶ potassium hydroxide + Ba3(PO4)2

Equilibrium constant

Construct the equilibrium constant, K, expression for: Ba(OH)_2 + K3PO4 ⟶ KOH + Ba3(PO4)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 Ba(OH)_2 + 2 K3PO4 ⟶ 6 KOH + Ba3(PO4)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 Ba(OH)_2 | 3 | -3 K3PO4 | 2 | -2 KOH | 6 | 6 Ba3(PO4)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) K3PO4 | 2 | -2 | ([K3PO4])^(-2) KOH | 6 | 6 | ([KOH])^6 Ba3(PO4)2 | 1 | 1 | [Ba3(PO4)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 = ([Ba(OH)2])^(-3) ([K3PO4])^(-2) ([KOH])^6 [Ba3(PO4)2] = (([KOH])^6 [Ba3(PO4)2])/(([Ba(OH)2])^3 ([K3PO4])^2)
Construct the equilibrium constant, K, expression for: Ba(OH)_2 + K3PO4 ⟶ KOH + Ba3(PO4)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 Ba(OH)_2 + 2 K3PO4 ⟶ 6 KOH + Ba3(PO4)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 Ba(OH)_2 | 3 | -3 K3PO4 | 2 | -2 KOH | 6 | 6 Ba3(PO4)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) K3PO4 | 2 | -2 | ([K3PO4])^(-2) KOH | 6 | 6 | ([KOH])^6 Ba3(PO4)2 | 1 | 1 | [Ba3(PO4)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 = ([Ba(OH)2])^(-3) ([K3PO4])^(-2) ([KOH])^6 [Ba3(PO4)2] = (([KOH])^6 [Ba3(PO4)2])/(([Ba(OH)2])^3 ([K3PO4])^2)

Rate of reaction

Construct the rate of reaction expression for: Ba(OH)_2 + K3PO4 ⟶ KOH + Ba3(PO4)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 Ba(OH)_2 + 2 K3PO4 ⟶ 6 KOH + Ba3(PO4)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 Ba(OH)_2 | 3 | -3 K3PO4 | 2 | -2 KOH | 6 | 6 Ba3(PO4)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 Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) K3PO4 | 2 | -2 | -1/2 (Δ[K3PO4])/(Δt) KOH | 6 | 6 | 1/6 (Δ[KOH])/(Δt) Ba3(PO4)2 | 1 | 1 | (Δ[Ba3(PO4)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[Ba(OH)2])/(Δt) = -1/2 (Δ[K3PO4])/(Δt) = 1/6 (Δ[KOH])/(Δt) = (Δ[Ba3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Ba(OH)_2 + K3PO4 ⟶ KOH + Ba3(PO4)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 Ba(OH)_2 + 2 K3PO4 ⟶ 6 KOH + Ba3(PO4)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 Ba(OH)_2 | 3 | -3 K3PO4 | 2 | -2 KOH | 6 | 6 Ba3(PO4)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 Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) K3PO4 | 2 | -2 | -1/2 (Δ[K3PO4])/(Δt) KOH | 6 | 6 | 1/6 (Δ[KOH])/(Δt) Ba3(PO4)2 | 1 | 1 | (Δ[Ba3(PO4)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[Ba(OH)2])/(Δt) = -1/2 (Δ[K3PO4])/(Δt) = 1/6 (Δ[KOH])/(Δt) = (Δ[Ba3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| barium hydroxide | K3PO4 | potassium hydroxide | Ba3(PO4)2 formula | Ba(OH)_2 | K3PO4 | KOH | Ba3(PO4)2 Hill formula | BaH_2O_2 | K3O4P | HKO | Ba3O8P2 name | barium hydroxide | | potassium hydroxide | IUPAC name | barium(+2) cation dihydroxide | | potassium hydroxide |
| barium hydroxide | K3PO4 | potassium hydroxide | Ba3(PO4)2 formula | Ba(OH)_2 | K3PO4 | KOH | Ba3(PO4)2 Hill formula | BaH_2O_2 | K3O4P | HKO | Ba3O8P2 name | barium hydroxide | | potassium hydroxide | IUPAC name | barium(+2) cation dihydroxide | | potassium hydroxide |

Substance properties

| barium hydroxide | K3PO4 | potassium hydroxide | Ba3(PO4)2 molar mass | 171.34 g/mol | 212.26 g/mol | 56.105 g/mol | 601.92 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 300 °C | | 406 °C | boiling point | | | 1327 °C | density | 2.2 g/cm^3 | | 2.044 g/cm^3 | solubility in water | | | soluble | dynamic viscosity | | | 0.001 Pa s (at 550 °C) |
| barium hydroxide | K3PO4 | potassium hydroxide | Ba3(PO4)2 molar mass | 171.34 g/mol | 212.26 g/mol | 56.105 g/mol | 601.92 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 300 °C | | 406 °C | boiling point | | | 1327 °C | density | 2.2 g/cm^3 | | 2.044 g/cm^3 | solubility in water | | | soluble | dynamic viscosity | | | 0.001 Pa s (at 550 °C) |

Units

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