15 Facts on HI + NaNO2: What, How To Balance & FAQs

Organic reactions produce compounds having immense utility in several organometallic catalysis process and organic synthesis. Let us explore the chemical reactivity of HI + NaNO2.

HI functions as a reducing agent used in the reduction reaction of aromatic nitro compounds into anilines. It is also used as a co catalyst in the Cativa process for the production of acetic acid from methanol. NaNO2 is a hygroscopic powder used as an antimicrobial product.

The reactivity of HI and NaNO2 can be a source of various important compounds utilized as food additives or pesticides. Thus, these important reaction mechanisms are studied in detail:

What is the product of HI and NaNO2?

HI and NaNO2 interact to produce sodium iodide (NaI), nitric oxide (NO), iodine gas (I2), and water (H2O). The complete chemical equation is given as:

HI + NaNO2 = NaI + NO + H2O + I2

What type of reaction is HI + NaNO2?

HI + NaNO2 is a substitution reaction followed by neutralization reaction where hydrogen in hydrogen iodide is substituted by sodium and later water is released along with a salt, NaI in an acid base neutralization process.

How to balance HI + NaNO2?

The following algebraic methodology can be used as an explanation to balance the chemical reaction

NaNO2 + HI = NaI + NO + H2O + I2,

  • Label each species (reactant or product) given in the chemical equation with a respective variable (A, B, C, D, E, and F) to illustrate unknown coefficients.
  • A NaNO2 + B HI = C I2 + D NO + E NaI + F H2O
  • Deduce a suitable equation for each element in the reacting species representing the number of atoms of an element in each reactant or product species, applied to solve the equation.
  • Na = A = E, N = A = D, O = 2A = D + F, H = B = F, I = B = C + E
  • The Gaussian elimination and substitution methodology is applied to simplify all the variables and coefficients, and the outcomes are
  • A = 2, B = 4, C = 1, D = 2, E = 2, and F = 2
  • Hence, the overall balanced equation is,
  • 2 NaNO2 + 4 HI = I2 + 2 NO + 2 NaI + 2 H2O

HI + NaNO2 titration

HI + NaNO2 system is performed as diazotization titration under cold conditions. Aromatic primary amine reacts with NaNO2 in an acidic solution of HI at 15 ℃ to form diazonium salt. The following steps are followed to proceed with the titration:

Apparatus and Chemicals Used

Burette, holder, stands, conical flask, volumetric flask, beakers, measuring cylinder, sulphanilamide, hydroiodic acid, sodium nitrate, distilled water

Indicator

Starch is used as an external indicator for HI + NaNO2 titration system

Procedure

  • 0.1 M sodium nitrate solution is prepared by dissolving 7.5 g of it in 1000 ml of a volumetric flask.
  • 0.5 g of sulphanilamide is transferred to a beaker.
  • To the beaker, 20 ml of HI and 50 ml of distilled water are added
  • The mixture is stirred until dissolution and cooled to 15 ℃ in an ice bath.
  • The titrated solution readily produces a distinct blue ring as touched by starch paper.
  • The titration is completed as the end point is observed.
  • Unknown concentration of sulphanilamide is calculated using the formula M1 * V= M2 * V2
  • Where M1 = Molarity of HI solution, V1 = Volume of HI solution, M2 = Molarity of sodium nitrate solution, V2 = Volume of sodium nitrate solution  

HI + NaNO2 net ionic equation

The net ionic equation of HI + NaNO2 is

NaNO2 (s) + H+ + I = NaI (s) + NO (g) + H+ + OH + I2 (g)

The net ionic equation is obtained using the following steps

  • Write the balanced chemical equation and designate the physical states of reactants and products accordingly
  • NaNO2 (s) + HI (l) = NaI (s) + NO (g) + H2O (l) + I2 (g)
  • Now, strong acids, bases, and salts dissociate into ions whereas pure solid substances and molecules do not dissociate
  • Thus, the net ionic equation is
  • NaNO2 (s) + H+ + I = NaI (s) + NO (g) + H+ + OH + I2 (g)

HI + NaNO2 conjugate pairs

  • Conjugate pair of strong acid HI is I.
  • NaNO2 does not form a conjugate pair as there is no free proton available for donation or acceptance.  

HI + NaNO2 intermolecular forces

Intermolecular forces acting on HI and NaNO2 are:

  • HI interacts using strong hydrogen bonding, weak London dispersion forces, and dipole-dipole interactions among the molecules.
  • NaNO2 being an ionic compound interacts forming ion-dipole intermolecular bonding between Na+ and NO2-2 ions.

HI + NaNO2 reaction enthalpy

HI + NaNO2 exhibits a negative reaction enthalpy of -20.1 kJ/mol. Enthalpy information for the reactants and products involved are as follows:

  • Enthalpy of formation for reactant HI: +25.95 kJ/mol
  • Enthalpy of formation for reactant NaNO2: -427 kJ/mol
  • Enthalpy of formation for product NaI: -288 kJ/mol
  • Enthalpy of formation for product H2O: -285.8 kJ/mol
  • Enthalpy of formation for product I2: +62.4 kJ/mol
  • Enthalpy of formation for product NO: +90.25 kJ/mol

Is HI + NaNO2 a buffer solution?

HI + NaNO2 does not form a buffer because the buffer is prepared by the addition of weak acid to the salt of its conjugate base whereas HI is not a weak acid and NaNO2 is not a salt of the conjugate base of HI.

Is HI + NaNO2 a complete reaction?

HI + NaNO2 is a complete reaction as the stable products namely NaI, water, NO, and I2 gas are produced in the reaction.   

Is HI + NaNO2 an exothermic or endothermic reaction?

HI + NaNO2 is an exothermic reaction because the calculated reaction enthalpy was noted to be negative which means heat is liberated in the process.

Is HI + NaNO2 a redox reaction?

HI + NaNO2 is not a redox reaction because hydrogen and sodium do not change and remain at +1 the oxidation state at both sides of the reactant and product in the ongoing reaction.

nano2
Representation of reaction with respect to oxidation state

Is HI + NaNO2 a precipitation reaction?

HI + NaNO2 is not a precipitation reaction as NaI produced in the reaction is highly soluble in water thus no solid precipitate remains when the reaction ends.

Is HI + NaNO2 reversible or irreversible reaction?

HI + NaNO2 is an irreversible reaction because the products formed in the reaction do not revert back to change into original reactants until the conditions are kept the same.

Is HI + NaNO2 displacement reaction?

HI + NaNO2 is a displacement reaction because the hydrogen cation is replaced by the sodium cation to form the new ionic compound NaI in the reaction.

Conclusions

The chemical reaction of HI + NaNO2 forms nitric oxide and sodium iodide as the major compound along with the liberation of iodine gas and water. NO is of huge industrial importance being a signaling molecule in physiological processes. NaI is primarily used in scintillation detectors.