15 Facts on HI + Na2S: What, How to Balance & FAQs

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Hydrogen iodide(HI) is generally a colorless gas that converts into a strong acid on dissolving in water. Let us see the reactivity of HI with Na2S through this article.

Sodium sulfide (Na2S) is a salt that can exist in both anhydrous and hydrous form. It is a highly reactive alkaline so it rapidly reacts with an acid to produce hydrogen sulfide. When Na2S is exposed to moist, the toxic fumes of hydrogen sulfide are emitted through its hydrides.

In this article, we will discuss important facts and FAQs about HI + Na2S chemical reactions such as reaction enthalpy, the heat required, the product formed, the type of reaction, the type of intermolecular forces between their compounds, etc.

What is the product of HI and Na2S?

Hydrogen iodide(HI) reacts with sodium sulfide(Na2S) to produce hydrogen sulfide(H2S) and sodium iodide(NaI). The chemical equation for the reaction is as follows:

2HI + Na2S   = H2S + 2NaI

What type of reaction is HI + Na2S?

HI + Na2S is a double displacement reaction but also follows the acid-base reaction because Na2S accepts a proton from HI and unshared electron pairs present on sulfur attract a proton rather than staying connected with the sodium ion. Thus, HI act as Bronsted acid, and Na2S act as Bronsted base.

How to balance HI + Na2S?

The overall balanced molecular equation for HI + Na2S is,

2HI + Na2S = H2S + 2NaI

The balanced equation for HI + Na2S can be derived by using the following method:

  • The unbalanced chemical equation for HI + Na2S,
    HI + Na2S = H2S + NaI
  • The number of all atoms in L.H.S and R.H.S of the chemical reaction should be equal.
  • If all atoms are not equal in number, some coefficient should be placed in front of the unbalanced molecule to increase the number of that particular unbalanced atom.  
  • Here we can see, the number of hydrogen and sodium atom is not the same on both sides of the reaction.
  • Therefore, to balance them, HI on the L.H.S. and NaI on R.H.S. is multiplied by the coefficient of 2.
  • Hence, the balanced chemical equation is,
  • 2HI + Na2S  = H2S + 2NaI

HI + Na2S titration

Quantitative analysis of HI can not be estimated by performing the titration of HI against Na2S, because HI is a strong acid and Na2S acts as a salt. So titration of this reaction is not possible. 

HI + Na2S net ionic equation

The overall balanced molecular equation of HI + Na2S is as follows:

2HI + Na2S = H2S + 2NaI

The net ionic equation for this chemical equation can be derived by following steps:

  • Write the solubility equation for HI + Na2S
  • Label the state or phase (s, l, g or aq) of each substance in the balanced molecular equation of HI + Na2S,
    2HI (aq) + Na2S(aq) = H2S(s) + 2NaI(aq)
  • Break all aquatic soluble ionic substances into their corresponding ions to get the complete ionic equation,
    2H+(aq) +2I–(aq) + 2Na+ (aq) + S2- (aq) = H2S (s) + 2Na+ (aq) + 2I–(aq)
  • To get the net ionic equation, remove spectator ions (I– and Na+) from L.H.S and R.H.S of the complete ionic equation.
  • Thus, the net ionic equation for HI + Na2S is,
    2H+(aq) + S2- (aq) = H2S (s)

 HI + Na2S conjugate pairs

Conjugate acid or base in 2HI + Na2S = H2S + 2NaI are:

  • HI (Acid) + H2O = H3O+ + I‑ (Conjugate Base)
  • Na2S does not contain any conjugate pair because it is a salt.
  • H2S (Acid) + H2O = H+ + HS‑ (Conjugate Base)
  • NaI does not contain any conjugate pair because it is a salt.

HI and Na2S intermolecular forces

Each compound in HI + Na2S = H2S + 2NaI contain the following inter molecular forces:

  • HI is a polar molecule so it contains ionic interaction, dipole – dipole interaction, and hydrogen bonding between its molecules.
  • Na2S is an ionic compound so it contains electrostatic attraction in its molecules. The positive charge on the sodium ion of one molecule bind with the negatively charged sulfide ion of another molecule
  • Bonding in the atoms of the H2S molecule is highly polar due to the high electronegativity difference between hydrogen and sulfur atoms. Due to this H2S shows strong dipole- dipole interaction and London dispersion force.  
  •  NaI is an ionic molecule, so its ions held each other with electrostatic and coulombic force.

HI + Na2S reaction enthalpy

HI + Na2S reaction enthalpy is 237.21 kJ/mol. The standard enthalpy of the formation of reactants and products is as follows:

Compound Standard Formation Enthalpy (ΔfH°(Kj/mol) )
HI26.46
Na2S -369.0
H2S -20.27
NaI-266.51
Standard Formation Enthalpy of Compounds

ΔHf = Sum of enthalpy of formation (products) – Sum enthalpy of formation (reactants)

ΔHf = [2*(26.46) – 369.0) – (-20.27 – 2*(266.51))]

ΔHf = [-316.08- (-553.29)]

ΔHf = 237.21 kJ/mol

Is HI + Na2S a buffer solution ?

HI + Na2S is not a buffer solution because the aqueous solution of this mixture does not provide corresponding acid or base.

Is HI + Na2S a complete reaction?

HI + Na2S is a complete reaction because in this reaction all molecules of HI and Na2S are consumed and converted into the final product successfully.

Is HI + Na2S an exothermic or endothermic reaction?

An endothermic reaction occurs when two moles of HI react with one mole of Na2S.

  • Enthalpy of reaction was found positive (i.e., ΔHf>0) from the above calculations.
  • Reactants absorb net 237.21 kJ/mol heat during the reaction which lowers the energy of the surroundings and makes the reaction system cool.

Is HI+ Na2S a redox reaction?

HI + Na2S is not a redox reaction because the oxidation states of elements do not change during the reaction.

Is HI + Na2S a precipitation reaction?

HI+ Na2S reaction is a precipitation reaction because H2S is produced in the solid phase after the completion of the reaction.

Is HI + Na2S reversible or irreversible reaction?

HI + Na2S is an irreversible reaction. Because it occurs only in the forward direction and produces H2S but it is not possible to get back Na2S at the ordinary reaction temperature, pressure, and reaction media.

Is HI + Na2S displacement reaction?

HI + Na2Sis a double displacement reaction because in this reaction, iodine ion (I–) and Sulphur ion (S2-) exchange their places with each other to form new products H2S and NaI.

Conclusion

The article states that Na2S reacts with strong acid HI and successfully forms NaI and a weak acid (H2S). H2S are obtained in the solid state, so it gets precipitated in reaction system and easily separated out. This reaction is endothermic as the net energy of reactants is higher than that of end products.