HCl, also known as hydrochloric acid, is a strong inorganic acid, and KBr (potassium bromide) is a salt. Let us discuss the reaction between HCl and KBr in detail.
Hydrochloric acid has characteristic pungent smell and has wide applications in metal refining process. Potassium Bromide is an ionic salt (formed by the reaction between KOH and HBr) that when combined with hydrochloric acid gives alkali metal salt and an acid.
Here, we will discuss the reaction products, enthalpy, and buffer solution with detailed explanations.
What is the product of HCl and KBr?
HCl reacts with KBr to give potassium chloride (KCl) and hydrogen bromide (HBr). The reaction is as follows;
HCl(aq) + KBr(aq) → KCl(aq) + HBr(aq)
What type of reaction is HCl + KBr
The reaction between HCl and KBr is a double-displacement reaction. Potassium being more reactive than hydrogen displaces it and forms KCl. Furthermore, the hydron (H+) combines with the bromide ion present in the solution to form HBr.
How to balance HCl + KBr
- To balance the equation (HCl +KBr) first we need to equate the number of atoms on either side of the equation.
- Count the number of atoms like H, Cl, K, and Br and simply add stoichiometry coefficients before the compounds to get the balanced equation. Here in this case, number of all the atoms in reactant and product side is equal.
- Thus the balanced equation will be-
HCl(aq) + KBr(aq) → KCl(aq) + HBr(aq)
HCl + KBr titration
KBr will not be titrated against HCl as no distinct color change will be observed even when indicator like Phenolphthalein is used, because it is colourless in acidic condition.
HCl + KBr net ionic equation
To obtain the net ionic equation for the reaction of HCl with KBr, we need to proceed in steps;
- Write the balanced molecular equation as–
HCl + KBr → KCl + HBr
- The physical states or phases (solid, liquid, gas, aqueous) of all the reactants and products are mentioned. Here all of the reactants and products are in the aqueous phase. The equation is now written as–
HCl(aq) + KBr(aq) → KCl(aq) + HBr(aq)
- Dissociate the electrolytes into ions. This can be written as–
H+Cl–(aq) + K+Br–(aq) → K+Cl–(aq) + H+Br–(aq)
- Balance the ions on either side of the equation. Although, here the number of ions is already balanced so crossing them out is not required. Hence, the net ionic equation becomes–
HCl(aq) + KBr(aq) → KCl(aq) + HBr(aq)
HCl + KBr conjugate pairs
- HCl gives H+ to form its conjugate base and Br– behaves as a base and takes up H+ to form its conjugate acid HBr.
HCl and KBr intermolecular forces
- The intermolecular forces in HCl are dipole-dipole interactions and London dispersion forces.
- KBr is an ionic compound so it will show ion-dipole interactions.
HCl + KBr reaction enthalpy
The enthalpy of the reaction between HCl and KBr is 87.98 KJ, which is positive.
Is HCl + KBr a buffer solution
HCl+KBr will not function as a buffer solution. This is because HCl is a strong acid rather than a weak acid, and also because the salt obtained will not constitute a buffer solution.
Is HCl + KBr a complete reaction
The reaction between HCl and KBr is a complete displacement reaction, as the product obtained is completely soluble and will not undergo further reaction.
Is HCl + KBr an exothermic or endothermic reaction
The reaction between HCl and KBr is an endothermic reaction as heat is absorbed during the reaction.
Is HCl + KBr a redox reaction
The reaction between HCl and KBr is not a redox reaction as there is no change in oxidation state observed for any of the elements.
Is HCl + KBr a precipitation reaction
The reaction HCl + KBr is not a precipitation reaction as it does not give any precipitate. The salt KCl obtained and the acid HBr is soluble in water.
Is HCl + KBr reversible or irreversible reaction
The reaction between HCl and KBr is reversible. This is because the reaction is more feasible if it goes from more acidic or basic compounds to less acidic or basic compounds. HBr is more acidic than HCl, so the backward reaction will be more favorable.
Is HCl + KBr displacement reaction
The hydrochloric acid reacts with potassium bromide via displacement reaction. This is an example of a double displacement reaction.
How to balance KBrO3 + KBr + HCl = Br2 + H2O + KCl
- We need to first indicate separately the oxidation states of the elements undergoing oxidation and reduction respectively. Here, the bromine atom (Br) in KBr is getting oxidized from -1 oxidation state to zero oxidation state, as well as reduced from +5 oxidation state to zero oxidation state.
1. Oxidation: Br-1 → Br0 + e–
2. Reduction: Br5+ + 5e– → Br0
- To equate the number of electrons, we will multiply the equation 1 (oxidation) by 5 and get-
3. 5KBr-1 → 5Br0 + 5e–
- On adding equations 2 and 3 we will get-
KBrO3 + 5KBr + HCl → 3Br2 + H2O + KCl
- Now, balance the number of other elements H, O, and Cl by adding the coefficients and we will get the balanced equation as-
KBrO3 + 5KBr + 6HCl → 3Br2 + 3H2O + 6KCl
How to balance Na2HAsO3 + KBrO3 + HCl = NaCl + KBr + H3AsO4
- The equation is divided into two halves–
1. Oxidation: As+3 → As+5 + 2e–
2. Reduction: Br+5 + 6e– → Br-1
- Equation 1 will be multiplied by 3 to get 6e– on either side. The equation will become–
3. Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + 3H3AsO4
- Further, the elements like H, Cl, K, and Br are balanced by simply adding the coefficients.
- Thus the balanced equation is given by-
3Na2HAsO3 + KBrO3 + 6HCl → 6NaCl + KBr + 3H3AsO4
How to balance KI + KBrO3 + HCl = I2 + KBr + KCl + H2O
- Elements undergoing reduction and oxidation are depicted below-
1. Oxidation: 2I– → I20 + 2e–
2. Reduction: Br+5 + 6e– → Br-1
- To equalize the charge on either side, equation 1 is multiplied by 3. The resulting equation becomes-
3KI + KBrO3 + HCl → 3I2 + KBr + KCl + H2O
- The other elements were balanced using stoichiometric coefficients.
- Thus the balanced equation becomes-
6KI + KBrO3 + 6HCl → 3I2 + KBr + 6KCl + 3H2O
Conclusion
The HCl-KBr reaction is an endothermic, double-displacement reaction. The products obtained from the reaction, KCl and HBr, are widely used as fertilizers and brominating agents, respectively.
Hi! I am Lubna Khan. I have done my Postgraduation in Chemistry at Jamia Millia Islamia, New Delhi. I have been in academia for years and have always welcomed new opportunities, lifestyles, and cultures coming my way.
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