NaOH + H3BO3 = H2O + Na2B4O7

NaOH (sodium hydroxide) + B(OH)_3 (boric acid) ⟶ H_2O (water) + Na_2B_4O_7 (sodium tetraborate)

Balance the chemical equation algebraically: NaOH + B(OH)_3 ⟶ H_2O + Na_2B_4O_7 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 B(OH)_3 ⟶ c_3 H_2O + c_4 Na_2B_4O_7 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and B: H: | c_1 + 3 c_2 = 2 c_3 Na: | 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 NaOH + 4 B(OH)_3 ⟶ 7 H_2O + Na_2B_4O_7

+ ⟶ +

sodium hydroxide + boric acid ⟶ water + sodium tetraborate

| sodium hydroxide | boric acid | water | sodium tetraborate molecular enthalpy | -425.8 kJ/mol | -1094 kJ/mol | -285.8 kJ/mol | -3291 kJ/mol total enthalpy | -851.6 kJ/mol | -4377 kJ/mol | -2001 kJ/mol | -3291 kJ/mol | H_initial = -5229 kJ/mol | | H_final = -5292 kJ/mol | ΔH_rxn^0 | -5292 kJ/mol - -5229 kJ/mol = -63.11 kJ/mol (exothermic) | | |

| sodium hydroxide | boric acid | water | sodium tetraborate molecular free energy | -379.7 kJ/mol | -968.9 kJ/mol | -237.1 kJ/mol | -3096 kJ/mol total free energy | -759.4 kJ/mol | -3876 kJ/mol | -1660 kJ/mol | -3096 kJ/mol | G_initial = -4635 kJ/mol | | G_final = -4756 kJ/mol | ΔG_rxn^0 | -4756 kJ/mol - -4635 kJ/mol = -120.7 kJ/mol (exergonic) | | |

Construct the equilibrium constant, K, expression for: NaOH + B(OH)_3 ⟶ H_2O + Na_2B_4O_7 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 NaOH + 4 B(OH)_3 ⟶ 7 H_2O + Na_2B_4O_7 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 NaOH | 2 | -2 B(OH)_3 | 4 | -4 H_2O | 7 | 7 Na_2B_4O_7 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) B(OH)_3 | 4 | -4 | ([B(OH)3])^(-4) H_2O | 7 | 7 | ([H2O])^7 Na_2B_4O_7 | 1 | 1 | [Na2B4O7] 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 = ([NaOH])^(-2) ([B(OH)3])^(-4) ([H2O])^7 [Na2B4O7] = (([H2O])^7 [Na2B4O7])/(([NaOH])^2 ([B(OH)3])^4)

Construct the rate of reaction expression for: NaOH + B(OH)_3 ⟶ H_2O + Na_2B_4O_7 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 NaOH + 4 B(OH)_3 ⟶ 7 H_2O + Na_2B_4O_7 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 NaOH | 2 | -2 B(OH)_3 | 4 | -4 H_2O | 7 | 7 Na_2B_4O_7 | 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) B(OH)_3 | 4 | -4 | -1/4 (Δ[B(OH)3])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) Na_2B_4O_7 | 1 | 1 | (Δ[Na2B4O7])/(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[NaOH])/(Δt) = -1/4 (Δ[B(OH)3])/(Δt) = 1/7 (Δ[H2O])/(Δt) = (Δ[Na2B4O7])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | boric acid | water | sodium tetraborate formula | NaOH | B(OH)_3 | H_2O | Na_2B_4O_7 Hill formula | HNaO | BH_3O_3 | H_2O | B_4Na_2O_7 name | sodium hydroxide | boric acid | water | sodium tetraborate IUPAC name | sodium hydroxide | boric acid | water | disodium 3, 7-dioxido-2, 4, 6, 8, 9-pentaoxa-1, 3, 5, 7-tetraborabicyclo[3.3.1]nonane

| sodium hydroxide | boric acid | water | sodium tetraborate molar mass | 39.997 g/mol | 61.83 g/mol | 18.015 g/mol | 201.21 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | 323 °C | 160 °C | 0 °C | 741 °C boiling point | 1390 °C | | 99.9839 °C | density | 2.13 g/cm^3 | | 1 g/cm^3 | 2.367 g/cm^3 solubility in water | soluble | | | surface tension | 0.07435 N/m | | 0.0728 N/m | dynamic viscosity | 0.004 Pa s (at 350 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless | odorless