HNO3 + KOH = KNO3H2O

HNO_3 nitric acid + KOH potassium hydroxide ⟶ KNO3H2O

Balance the chemical equation algebraically: HNO_3 + KOH ⟶ KNO3H2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 KOH ⟶ c_3 KNO3H2O Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and K: H: | c_1 + c_2 = 2 c_3 N: | c_1 = c_3 O: | 3 c_1 + c_2 = 4 c_3 K: | c_2 = c_3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | HNO_3 + KOH ⟶ KNO3H2O

+ ⟶ KNO3H2O

nitric acid + potassium hydroxide ⟶ KNO3H2O

Construct the equilibrium constant, K, expression for: HNO_3 + KOH ⟶ KNO3H2O 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: HNO_3 + KOH ⟶ KNO3H2O 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 HNO_3 | 1 | -1 KOH | 1 | -1 KNO3H2O | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 1 | -1 | ([HNO3])^(-1) KOH | 1 | -1 | ([KOH])^(-1) KNO3H2O | 1 | 1 | [KNO3H2O] 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 = ([HNO3])^(-1) ([KOH])^(-1) [KNO3H2O] = ([KNO3H2O])/([HNO3] [KOH])

Construct the rate of reaction expression for: HNO_3 + KOH ⟶ KNO3H2O 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: HNO_3 + KOH ⟶ KNO3H2O 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 HNO_3 | 1 | -1 KOH | 1 | -1 KNO3H2O | 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 HNO_3 | 1 | -1 | -(Δ[HNO3])/(Δt) KOH | 1 | -1 | -(Δ[KOH])/(Δt) KNO3H2O | 1 | 1 | (Δ[KNO3H2O])/(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 = -(Δ[HNO3])/(Δt) = -(Δ[KOH])/(Δt) = (Δ[KNO3H2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | potassium hydroxide | KNO3H2O formula | HNO_3 | KOH | KNO3H2O Hill formula | HNO_3 | HKO | H2KNO4 name | nitric acid | potassium hydroxide |

| nitric acid | potassium hydroxide | KNO3H2O molar mass | 63.012 g/mol | 56.105 g/mol | 119.12 g/mol phase | liquid (at STP) | solid (at STP) | melting point | -41.6 °C | 406 °C | boiling point | 83 °C | 1327 °C | density | 1.5129 g/cm^3 | 2.044 g/cm^3 | solubility in water | miscible | soluble | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) |