KOH + Bi(NO3)3 = KNO3 + BiNO3(OH)2

KOH potassium hydroxide + Bi(NO3)3 ⟶ KNO_3 potassium nitrate + BiNO3(OH)2

Balance the chemical equation algebraically: KOH + Bi(NO3)3 ⟶ KNO_3 + BiNO3(OH)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Bi(NO3)3 ⟶ c_3 KNO_3 + c_4 BiNO3(OH)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Bi and N: H: | c_1 = 2 c_4 K: | c_1 = c_3 O: | c_1 + 9 c_2 = 3 c_3 + 5 c_4 Bi: | c_2 = c_4 N: | 3 c_2 = c_3 + 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 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + Bi(NO3)3 ⟶ 2 KNO_3 + BiNO3(OH)2

+ Bi(NO3)3 ⟶ + BiNO3(OH)2

potassium hydroxide + Bi(NO3)3 ⟶ potassium nitrate + BiNO3(OH)2

Construct the equilibrium constant, K, expression for: KOH + Bi(NO3)3 ⟶ KNO_3 + BiNO3(OH)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: 2 KOH + Bi(NO3)3 ⟶ 2 KNO_3 + BiNO3(OH)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 KOH | 2 | -2 Bi(NO3)3 | 1 | -1 KNO_3 | 2 | 2 BiNO3(OH)2 | 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) Bi(NO3)3 | 1 | -1 | ([Bi(NO3)3])^(-1) KNO_3 | 2 | 2 | ([KNO3])^2 BiNO3(OH)2 | 1 | 1 | [BiNO3(OH)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 = ([KOH])^(-2) ([Bi(NO3)3])^(-1) ([KNO3])^2 [BiNO3(OH)2] = (([KNO3])^2 [BiNO3(OH)2])/(([KOH])^2 [Bi(NO3)3])

Construct the rate of reaction expression for: KOH + Bi(NO3)3 ⟶ KNO_3 + BiNO3(OH)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: 2 KOH + Bi(NO3)3 ⟶ 2 KNO_3 + BiNO3(OH)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 KOH | 2 | -2 Bi(NO3)3 | 1 | -1 KNO_3 | 2 | 2 BiNO3(OH)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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) Bi(NO3)3 | 1 | -1 | -(Δ[Bi(NO3)3])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(Δt) BiNO3(OH)2 | 1 | 1 | (Δ[BiNO3(OH)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 = -1/2 (Δ[KOH])/(Δt) = -(Δ[Bi(NO3)3])/(Δt) = 1/2 (Δ[KNO3])/(Δt) = (Δ[BiNO3(OH)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | Bi(NO3)3 | potassium nitrate | BiNO3(OH)2 formula | KOH | Bi(NO3)3 | KNO_3 | BiNO3(OH)2 Hill formula | HKO | BiN3O9 | KNO_3 | H2BiNO5 name | potassium hydroxide | | potassium nitrate |

| potassium hydroxide | Bi(NO3)3 | potassium nitrate | BiNO3(OH)2 molar mass | 56.105 g/mol | 394.99 g/mol | 101.1 g/mol | 305 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 406 °C | | 334 °C | boiling point | 1327 °C | | | density | 2.044 g/cm^3 | | | solubility in water | soluble | | soluble | dynamic viscosity | 0.001 Pa s (at 550 °C) | | | odor | | | odorless |