Input interpretation
![KOH potassium hydroxide + B(OH)_3 boric acid ⟶ H_2O water + K2B4O7](../image_source/a00df54d378e47625b52ad6e900eb85f.png)
KOH potassium hydroxide + B(OH)_3 boric acid ⟶ H_2O water + K2B4O7
Balanced equation
![Balance the chemical equation algebraically: KOH + B(OH)_3 ⟶ H_2O + K2B4O7 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 B(OH)_3 ⟶ c_3 H_2O + c_4 K2B4O7 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and B: H: | c_1 + 3 c_2 = 2 c_3 K: | c_1 = 2 c_4 O: | c_1 + 3 c_2 = c_3 + 7 c_4 B: | c_2 = 4 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 = 2 c_2 = 4 c_3 = 7 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 4 B(OH)_3 ⟶ 7 H_2O + K2B4O7](../image_source/4e2b9a3900ee33819a0bb6cdb486f9cd.png)
Balance the chemical equation algebraically: KOH + B(OH)_3 ⟶ H_2O + K2B4O7 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 B(OH)_3 ⟶ c_3 H_2O + c_4 K2B4O7 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and B: H: | c_1 + 3 c_2 = 2 c_3 K: | c_1 = 2 c_4 O: | c_1 + 3 c_2 = c_3 + 7 c_4 B: | c_2 = 4 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 = 2 c_2 = 4 c_3 = 7 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 4 B(OH)_3 ⟶ 7 H_2O + K2B4O7
Structures
![+ ⟶ + K2B4O7](../image_source/5754a7832dbd2f32bd4b25ca3aa1897d.png)
+ ⟶ + K2B4O7
Names
![potassium hydroxide + boric acid ⟶ water + K2B4O7](../image_source/2e88f42fa49458a3b7c08ecebc55602e.png)
potassium hydroxide + boric acid ⟶ water + K2B4O7
Equilibrium constant
![Construct the equilibrium constant, K, expression for: KOH + B(OH)_3 ⟶ H_2O + K2B4O7 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: 2 KOH + 4 B(OH)_3 ⟶ 7 H_2O + K2B4O7 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 | 2 | -2 B(OH)_3 | 4 | -4 H_2O | 7 | 7 K2B4O7 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) B(OH)_3 | 4 | -4 | ([B(OH)3])^(-4) H_2O | 7 | 7 | ([H2O])^7 K2B4O7 | 1 | 1 | [K2B4O7] 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])^(-2) ([B(OH)3])^(-4) ([H2O])^7 [K2B4O7] = (([H2O])^7 [K2B4O7])/(([KOH])^2 ([B(OH)3])^4)](../image_source/d372a51afdfb2f2e550e6b4dcba3e7da.png)
Construct the equilibrium constant, K, expression for: KOH + B(OH)_3 ⟶ H_2O + K2B4O7 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: 2 KOH + 4 B(OH)_3 ⟶ 7 H_2O + K2B4O7 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 | 2 | -2 B(OH)_3 | 4 | -4 H_2O | 7 | 7 K2B4O7 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) B(OH)_3 | 4 | -4 | ([B(OH)3])^(-4) H_2O | 7 | 7 | ([H2O])^7 K2B4O7 | 1 | 1 | [K2B4O7] 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])^(-2) ([B(OH)3])^(-4) ([H2O])^7 [K2B4O7] = (([H2O])^7 [K2B4O7])/(([KOH])^2 ([B(OH)3])^4)
Rate of reaction
![Construct the rate of reaction expression for: KOH + B(OH)_3 ⟶ H_2O + K2B4O7 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: 2 KOH + 4 B(OH)_3 ⟶ 7 H_2O + K2B4O7 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 | 2 | -2 B(OH)_3 | 4 | -4 H_2O | 7 | 7 K2B4O7 | 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 | 2 | -2 | -1/2 (Δ[KOH])/(Δt) B(OH)_3 | 4 | -4 | -1/4 (Δ[B(OH)3])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) K2B4O7 | 1 | 1 | (Δ[K2B4O7])/(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 = -1/2 (Δ[KOH])/(Δt) = -1/4 (Δ[B(OH)3])/(Δt) = 1/7 (Δ[H2O])/(Δt) = (Δ[K2B4O7])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/ddfd8b325222cc743f409295100bced5.png)
Construct the rate of reaction expression for: KOH + B(OH)_3 ⟶ H_2O + K2B4O7 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: 2 KOH + 4 B(OH)_3 ⟶ 7 H_2O + K2B4O7 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 | 2 | -2 B(OH)_3 | 4 | -4 H_2O | 7 | 7 K2B4O7 | 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 | 2 | -2 | -1/2 (Δ[KOH])/(Δt) B(OH)_3 | 4 | -4 | -1/4 (Δ[B(OH)3])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) K2B4O7 | 1 | 1 | (Δ[K2B4O7])/(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 = -1/2 (Δ[KOH])/(Δt) = -1/4 (Δ[B(OH)3])/(Δt) = 1/7 (Δ[H2O])/(Δt) = (Δ[K2B4O7])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| potassium hydroxide | boric acid | water | K2B4O7 formula | KOH | B(OH)_3 | H_2O | K2B4O7 Hill formula | HKO | BH_3O_3 | H_2O | B4K2O7 name | potassium hydroxide | boric acid | water |](../image_source/08edcefa55cf5dee9b2852a6c238e249.png)
| potassium hydroxide | boric acid | water | K2B4O7 formula | KOH | B(OH)_3 | H_2O | K2B4O7 Hill formula | HKO | BH_3O_3 | H_2O | B4K2O7 name | potassium hydroxide | boric acid | water |
Substance properties
![| potassium hydroxide | boric acid | water | K2B4O7 molar mass | 56.105 g/mol | 61.83 g/mol | 18.015 g/mol | 233.43 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | melting point | 406 °C | 160 °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 | odorless |](../image_source/28445d37508de0135f8930f26ad28306.png)
| potassium hydroxide | boric acid | water | K2B4O7 molar mass | 56.105 g/mol | 61.83 g/mol | 18.015 g/mol | 233.43 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | melting point | 406 °C | 160 °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 | odorless |
Units