HI and Fe_{2}(SO_{4})_{3} are the chemical representation of hydrogen iodide, and ferric sulfate respectively. Let us investigate the interactions between these two constituents.

**Hydroiodic acid abbreviated as HI is an extremely volatile pale gas while iron (III) sulfate abbreviated as Fe _{2}(SO_{4})_{3}, is a reddish-brown solid which is unsolvable in water. Both interact to form a bright yellow solution.**

This article will deliver a close understanding of all the critical characteristics of the reaction between HI and Fe_{2}(SO_{4})_{3}.

**What is the product of HI+Fe**_{2}(SO_{4})_{3}?

_{2}(SO

_{4})

_{3}?

**The end products of the interaction of HI and Fe _{2}(SO_{4})_{3} produce FeI_{2} (iron (II) iodide) and H_{2}SO_{4} (**

**sulfuric acid**

**).**

**2HI + Fe _{2}(SO_{4})_{3} → FeI_{2} + 2H_{2}SO_{4}**

**What type of reaction is HI + Fe**_{2}(SO_{4})_{3}?

_{2}(SO

_{4})

_{3}?

**HI and Fe _{2}(SO_{4})_{3} react with one another to produce lethal fumes and lie below displacement reaction.**

**How to balance HI + Fe**_{2}(SO_{4})_{3}?

_{2}(SO

_{4})

_{3}?

**The imbalanced reaction for the HI + Fe _{2}(SO_{4})_{3} is specified below**

**HI + Fe _{2}(SO_{4})_{3} → FeI_{2} + 2H_{2}SO_{4}**

**The initial basic step is identifying the quantity of participating atoms in reactants and products.**

**The number of atoms on the reactant side is 1H, 1I, 2Fe, 12O, and 3S while the number of atoms on the product side is 2H, 1S, 4O, 2I, and 1Fe.**

**Initialize by matching the hydrogen on the reactant and product sides of the equation:**

**2HI +Fe**_{2}(SO_{4})_{3}→ FeI_{2}+ H_{2}SO_{4}

**Now equalise the iron and the sulfur atoms respectively:**

**2HI + 3**→ 2**F****e**_{2}(SO_{4})_{3}**FeI**+ 6_{2}**H**_{2}SO_{4}

**The final balanced equation is listed underneath:**

**2HI + 3****F****e**_{2}(SO_{4})_{3}**→ 2**+ 6**FeI**_{2}**H**_{2}SO_{4}

**HI + ****F****e**_{2}(SO_{4})_{3} **titration**

**F****e**_{2}(SO_{4})_{3}**HI +Fe_{2}(SO_{4})_{3} **

**titration**

**is performed to calculate the concentration of F**

**in moles required to thoroughly react with HI.****e**_{2}(SO_{4})_{3}__Apparatus__

__Apparatus__

**Pipette, rubber tube with a syringe connected, different-sized beakers (250 ml-1, 100 ml-3), grease stand with pencil, burette clamp, and a 250 ml flask**

__Indicator__

__Indicator__

**Acid-base indicator ****Methyl orange**** or ****Phenolphthalein**** is utilized to recognize the titration’s end point.**

__Procedure__

__Procedure__

**To make a**solution, dissolve a known quantity of the salt in a volumetric flask.**F****e**_{2}(SO_{4})_{3}**In a different volumetric flask, make a solution of the titrant, such as Hydroiodic acid, and add an indicator dropwise to the solution.****Put the solution of****F****e**_{2}(SO_{4})_{3}**in an****Erlenmeyer flask****.****Using a burette, start the titration process by gradually adding the titrant.****Keep track of how much titrant was added at each interval.****Throughout the titration, stir the solution.****Following the achievement of the endpoint, the volume of titrant added is noted and the quantity of****F****e**_{2}(SO_{4})_{3}**can be estimated.****For better outcomes replicate the procedure 2-3 times.****Determine the volume of****F**in the initial solution by deducting the volume of titrant that was used.**e**_{2}(SO_{4})_{3}

**HI + ****F****e**_{2}(SO_{4})_{3} net ionic equation

**net ionic equation**

**F****e**_{2}(SO_{4})_{3}**The Net ionic equation for HI + Fe_{2}(SO_{4})_{3} is HI + Fe^{2+} –> FeI_{2} + H^{+}**

**It can be written following underneath listed steps:**

**Fe_{2}(SO_{4})_{3}** + HI –> 2FeI

_{2}+ 3H

_{2}SO

_{4}

**Disrupt the equation into its constituent ions:**

**2Fe**^{2+}+ 3SO4_{2}^{–}+ HI –> 2FeI_{2}+ 3H^{+ }+ 3SO_{4}^{2-}

**Recognize and eliminate the spectator ions:**

**SO**_{4}^{2- }is performing as spectator ions and can be eliminated from the whole equation.

**2Fe**^{2+}+ 3SO_{4}^{2-}+ HI –> 2FeI_{2 }+ 3H^{+}+ 3SO_{4}^{2-}

**Write the overall ionic equation:**

**HI + Fe**^{2+}–> FeI_{2}+ H^{+}

**HI + ****F****e**_{2}(SO_{4})_{3} conjugate pairs

**conjugate pairs**

**F****e**_{2}(SO_{4})_{3}**HI + Fe _{2}(SO_{4})_{3} has the following conjugate pairs, **

**The conjugate pair for HI is H**.^{+}(hydrogen ion)**The conjugate pair of Fe**_{2}(SO_{4})_{3}is Fe^{3+}(ferric ion).

**HI + ****Fe**_{2}(SO_{4})_{3} intermolecular forces

**intermolecular forces**

**Fe**_{2}(SO_{4})_{3}**HI + Fe _{2}(SO_{4})_{3} has the following intermolecular forces, **

**Hydrogen bonds****and dipole-dipole interactions are the intermolecular forces that hold the molecules of hydrogen iodide (HI) and iron (III) sulfate (****F**) together.**e**_{2}(SO_{4})_{3}

**HI + F****e**_{2}(SO_{4})_{3} reaction enthalpy

**reaction enthalpy**

**e**_{2}(SO_{4})_{3}**For the reaction of HI + Fe_{2}(SO_{4})_{3}, the reaction enthalpy value is -898.3 kJ/mole. **

**Reaction Enthalpy = (enthalpies of formation of all products) – (enthalpies of formation of all reactants)****ΔH = ΔH**_{f}(products) – ΔH_{f }(reactants)**Where ΔHf signifies the enthalpy of formation of the particular species.****The following equation gives the enthalpy of reaction for this reaction:****ΔH = (-817.4 kJ/mol + -817.4 kJ/mol) – (-80.8 kJ/mol + -1520.1 kJ/mol) = -898.3 kJ/mol****As a result, the reaction between HI and****F****e**_{2}(SO_{4})_{3}**has a negative enthalpy of -898.3 kJ/mol.**

**Is HI + ****F****e**_{2}(SO_{4})_{3} **a buffer solution?**

**F**

**e**_{2}(SO_{4})_{3}**HI + Fe_{2}(SO_{4})_{3} mixture is not a buffer.**

**Because HI +**

**lacks both a weak acid and its conjugate base.****F****e**_{2}(SO_{4})_{3}**Is HI + ****F****e**_{2}(SO_{4})_{3} a complete reaction?

**a complete reaction?**

**F****e**_{2}(SO_{4})_{3}**HI +Fe_{2}(SO_{4})_{3}**

**is not a complete reaction as the reactants and products are in continuous dynamic balance and can be transformed back and forth into one another.**

**Is HI + ****F****e**_{2}(SO_{4})_{3} **an exothermic or endothermic reaction?**

**F****e**_{2}(SO_{4})_{3}**The fact that the calculated reaction’s enthalpy was negative indicates that HI +Fe_{2}(SO_{4})_{3} is **

**exothermic**

**in nature and produces a significant amount of heat energy.**

**Is HI + F****e**_{2}(SO_{4})_{3} **a redox reaction?**

**e**_{2}(SO_{4})_{3}**HI +Fe_{2}(SO_{4})_{3} is referred to as a **

**redox process**

**as the reactants exchange electrons with one another.**

**While the hydrogen ions are being reduced, the iron ions in the reactants (Fe**

^{2+}and Fe^{3+}) are being oxidized (losing electrons) (gaining electrons). Iron iodide (FeI2) and sulfuric acid are the end products (H_{2}SO_{4}).**Is HI + ****F****e**_{2}(SO_{4})_{3} a precipitation reaction?

**a precipitation reaction?**

**F****e**_{2}(SO_{4})_{3}**HI +Fe_{2}(SO_{4})_{3} cannot be referred to as a **

**precipitation**

**reaction because HI and**

**are both soluble salts in this reaction, neither insoluble salt is created.****F****e**_{2}(SO_{4})_{3}**Is HI + F****e**_{2}(SO_{4})_{3} a reversible or irreversible reaction?

**a reversible or irreversible reaction?**

**e**_{2}(SO_{4})_{3}**HI + Fe_{2}(SO_{4})_{3} is referred to as a **

**reversible reaction**

**as the reaction shift back to the left after the formation of the products.**

**Is HI + **** ****F****e**_{2}(SO_{4})_{3} a displacement reaction?

**a displacement reaction?**

**F****e**_{2}(SO_{4})_{3}**HI + Fe_{2}(SO_{4})_{3} is referred to as a **

**displacement reaction**

**since iron replaces Sulphur in the molecule**

**in this reaction, forming FeI****F****e**_{2}(SO_{4})_{3}_{2}and H_{2}SO_{4}.**Conclusion**

To complete HI + Fe_{2}(SO_{4})_{3} is a displacement, a reversible reaction. The reaction on interaction forms deep yellow FeI_{2} (iron (II) iodide) and H_{2}SO_{4 }(sulfuric acid). It gives a negative reaction enthalpy, releases fumes, and hence is exothermic in nature.

Read more facts on HI

Hello…I’m Ritika Vaishnav, a Chemistry Subject Matter Expert with a passion for creating educational and engaging content on a wide range of topics related to chemistry. With a master’s degree in Chemistry, I have written for a variety of publications, including research papers. My writing is concise and well-researched, and I strive to make complex topics easily understandable for readers.

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