
Aqueous solution of $KCl$ is
(A)- acidic
(B)- basic
(C)- neutral
(D)- none of the above
Answer
499.2k+ views
Hint: During the hydrolysis process, the ions of the salt on dissolving in water change the concentration of the ${{H}^{+}}\text{or}\,O{{H}^{-}}$ in the solution, thus the acidity of the solution changes.
Complete step by step answer:
The given aqueous solution of potassium chloride is formed by the hydrolysis of the salt. This involves the salt being dissolved in water and breaking down into its constituent ions.
The ions present in potassium chloride is the potassium cation, ${{K}^{+}}$ and the chloride anion, $C{{l}^{-}}$. The salt obtained from the neutralization reaction of an acid with base, that is, hydrochloric acid with potassium hydroxide, on reaction gives potassium chloride and water.
Neutralization reaction:
\[HCl+KOH\to KCl+{{H}_{2}}O\]
The hydrochloric acid is a strong acid and the potassium hydroxide is a strong base because they dissociate completely and ionize in solution giving proton or hydroxide ion respectively as follows:
\[HCl\to {{H}^{+}}+Cl\]
\[KOH\to {{K}^{+}}+O{{H}^{-}}\]
Here, the $C{{l}^{-}}$is the conjugate base of strong acid, HCl and ${{K}^{+}}$ is the conjugate acid of strong base.
Therefore, the salt solution of this strong acid and strong base acts as an electrolyte with these ions present in its constituent forms. The nature of solution depends on the reactivity of the ions (that is, the conjugate base and conjugate acid) whether it produces either ${{H}^{+}}\,or\,O{{H}^{-}}$ ions with water.
But in this case, both the ions (${{K}^{+}}$and $C{{l}^{-}}$) do not react back with the ${{H}^{+}}\,or\,O{{H}^{-}}$ions of water, thus, concentrations of ions remain equal, that is, $\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]$, and the ${{K}^{+}}$and $C{{l}^{-}}$ ions do not hydrolyse. So, there is no effect on the acidity of the salt solution.
So, the correct answer is “Option C”.
Note: The aqueous KCl solution due to its neutral nature and complete ionization in solution, it is used for the calibration purpose in the experiment procedure for the electrical conductivity of ionic solutions.
Complete step by step answer:
The given aqueous solution of potassium chloride is formed by the hydrolysis of the salt. This involves the salt being dissolved in water and breaking down into its constituent ions.
The ions present in potassium chloride is the potassium cation, ${{K}^{+}}$ and the chloride anion, $C{{l}^{-}}$. The salt obtained from the neutralization reaction of an acid with base, that is, hydrochloric acid with potassium hydroxide, on reaction gives potassium chloride and water.
Neutralization reaction:
\[HCl+KOH\to KCl+{{H}_{2}}O\]
The hydrochloric acid is a strong acid and the potassium hydroxide is a strong base because they dissociate completely and ionize in solution giving proton or hydroxide ion respectively as follows:
\[HCl\to {{H}^{+}}+Cl\]
\[KOH\to {{K}^{+}}+O{{H}^{-}}\]
Here, the $C{{l}^{-}}$is the conjugate base of strong acid, HCl and ${{K}^{+}}$ is the conjugate acid of strong base.
Therefore, the salt solution of this strong acid and strong base acts as an electrolyte with these ions present in its constituent forms. The nature of solution depends on the reactivity of the ions (that is, the conjugate base and conjugate acid) whether it produces either ${{H}^{+}}\,or\,O{{H}^{-}}$ ions with water.
But in this case, both the ions (${{K}^{+}}$and $C{{l}^{-}}$) do not react back with the ${{H}^{+}}\,or\,O{{H}^{-}}$ions of water, thus, concentrations of ions remain equal, that is, $\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]$, and the ${{K}^{+}}$and $C{{l}^{-}}$ ions do not hydrolyse. So, there is no effect on the acidity of the salt solution.
So, the correct answer is “Option C”.
Note: The aqueous KCl solution due to its neutral nature and complete ionization in solution, it is used for the calibration purpose in the experiment procedure for the electrical conductivity of ionic solutions.
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