Hydrogen iodide(HI) is a yellow gas while copper sulfate (CuSO4), a blue crystalline solid liquifies in water to form copper sulfate solutions. Let us see some of their properties.
HI being very volatile reacts with numerous organic compounds. CuSO4 is a very vital compound in the industrial sector since it is utilized in an extensive variety of applications including agriculture, manufacturing, and construction.
This article will deliver thorough facts about the various characteristics of the interaction between HI and CuSO4.
What is the product of HI and CuSO4 ?
The product of the interaction between HI and CuSO4 is copper(I) iodide (CuI) along with sulfuric acid (H2SO4).
HI + CuSO4 → CuI + H2SO4
What type of reaction is HI and CuSO4
HI and CuSO4 lie under a double displacement reaction, also identified as a metathesis reaction.
How to balance HI and CuSO4?
HI + CuSO4 reaction is balanced in the following underneath steps,
- Count both the left and right sides of the equation to determine how many atoms there are for each element.
- Equalize the number of atoms for each element so that they are evenly distributed on both sides.
- For the elements HI and H2SO4, add coefficients (numbers in front of the element symbols) of 2 and 1, respectively, to bring them to the corresponding values on the right, which are 2 and 1.
- To make the CuSO4 and CuI elements’ respective coefficients on the right side equal to those on the left, add 1 and 2 there, respectively.
- Now that the equation is balanced:
- 2HI + CuSO4 → 2CuI + H2SO4
HI and CuSO4 titration
HI + CuSO4 titration is accomplished to determine the amount of HI in a solution. By identifying the concentration of CuSO4 utilized in the titration, the amount of HI available in the solution can be found.
Apparatus
Pipette, different-sized beakers (250 ml-1, 100 ml-3), grease stand with pencil, burette clamp, and a 250 ml flask
Indicator
The end-point of the titration is identified using the acid-base reagent Phenolphthalein.
Procedure
- Pour a volumetric flask with a known amount of HI solution.
- Fill the flask with a few drops of phenolphthalein indicator.
- Using a burette, add the CuSO4 solution to the flask.
- Stir the flask vigorously to achieve complete mixing.
- Pay attention to how the indicator’s color changes. The endpoint is attained when the solution takes on a faint pink hue.
- Note the quantity of CuSO4 solution that was used.
- Employ the titration equation(S1V1=S2V2) to determine the amount of HI present in the solution.
- For better outcomes replicate the procedure 2-3 times.
HI + CuSO4 net ionic equation
The net ionic equation for the reaction:
2H+ + Cu2+ + SO42- → 2Cu+ + H2SO4
- Dividing the chemical equation into its individual ions is the first step. This can be decomposed into
- 2H+ + I– + Cu2+ + SO42- → 2Cu+ + I– + H2SO4
- Equilibrating the equation is the next step. To accomplish this, multiply the reactants and products until they are all equal to one another. By multiplying the reactants and products by two, the equation in this instance can be made to balance.
- 2H+ + 2I– + 2Cu2+ + 2SO42- → 2Cu+ + 2I– + H2SO4 is the balanced equation.
- Eliminating the spectator ions is step three. The iodide ions are the spectator ions in this instance. By removing the iodide ions from the equation, we are left with
- 2H+ + Cu2+ + SO42- → 2Cu+ + H2SO4
- The net ionic equation must be written as the fourth and last step. The ions that are present on both the reactant and product sides of the equation are eliminated to achieve this. The net ionic equation in this instance is
- 2H+ + Cu2+ + SO42- → 2Cu+ + H2SO4
HI + CuSO4 conjugate pairs
HI + CuSO4 has the following conjugate pairs, differing by one proton:
- The conjugate acid-base pair of HI = H3O
- The conjugate acid-base pair of CuSO4 = Cu2+.
HI + CuSO4 intermolecular forces
HI + CuSO4 has the following intermolecular forces,
- Hydrogen bonds and interactions between dipoles are the intermolecular forces that hold HI and CuSO4 together.
- When the hydrogen atom of HI is drawn to the oxygen atom of CuSO4, hydrogen bonds are created. This is because the oxygen atom in CuSO4 is slightly negative and the hydrogen atom in HI is slightly positive.
- HI and CuSO4 also interact via dipole-dipole interactions. This is because HI is slightly polar, which means that its electrons are distributed unevenly, and CuSO4 is similarly polar, which means that its electrons are distributed unevenly.
HI + CuSO4 reaction enthalpy
The reaction enthalpy of HI + CuSO4 is -80.14 kJ/mol. This reaction can be stated as follows:
- HI (g) + CuSO4 (s) → CuI (s) + H2SO4 (g)
- The reaction enthalpy of this reaction can be considered using the subsequent equation:
- ΔH = ΣHf (products) – ΣHf (reactants)
- Where ΣHf is the sum of the standard enthalpies of the formation of the reactants and products.
- Therefore, the reaction enthalpy can be calculated as follows:
- ΔH = (-890.1 kJ/mol) + (-41.7 kJ/mol) – (-100.7 kJ/mol)
- ΔH = -80.14 kJ/mol.
Is HI + CuSO4 a buffer solution?
HI + CuSO4 cannot be referred to as a buffer solution. As the modest amounts of acid or base are added, the pH of a buffer solution does not change. This definition is not met by HI + CuSO4.
Is HI + CuSO4 a complete reaction?
HI + CuSO4 is not really a complete reaction since the reactants and products can change back and forth into one another in constant dynamic equilibrium.
Is HI + CuSO4 an exothermic or endothermic reaction?
HI + CuSO4 is exothermic and generates a sizable amount of heat energy.
Is HI + CuSO4 a redox reaction?
HI + CuSO4 is stated to as a redox process. CuSO4 undergoes oxidation in this process, to produce CuI. Reduced H2 gets electrons to create H2SO4.
Is HI + CuSO4 a precipitation reaction?
HI + CuSO4 is stated to as a precipitation reaction because two ions (Cu2+ and SO42-) in the reactants interact to produce a new molecule, a solid precipitate (CuI) is formed. The sulphate ions (SO42-) and hydronium ions (H3O+) are still in solution.
Is HI + CuSO4 a reversible or irreversible reaction?
HI + CuSO4 is stated to as a reversible reaction. As the left-side reactants (2HI and CuSO4) can become the products (2CuI and H2SO4), and the products (2CuI and H2SO4) can be transformed back into reactants through the reaction 2HI + CuSO4 = 2CuI + H2SO4 (2HI and CuSO4).
Is HI + CuSO4 a displacement reaction?
HI + CuSO4 is stated to as a displacement reaction. As iodine (I) in this reaction displaces copper (Cu) from copper sulphate (CuSO4) (I). Copper iodide (CuI) and sulfuric acid (H2SO4), are the byproducts of the process
Conclusion
To broad HI + CuSO4 is a reversible reaction which forms yellow colored precipitate resulting into release of energy. Interaction between both results into formation of displaced product copper iodide (CuI) and sulfuric acid (H2SO4) respectively.
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