Answer
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Hint: \[{K_a}\] is the acid dissociation constant and it is the equilibrium constant for the dissociation of acid.
Whenever the compound tends to readily lose its proton by dissociation, then it indicates that the compound is highly acidic. \[p{K_a}\] is the negative logarithm of the acid dissociation constant.
I.e.\[p{K_a} = - \log \left( {{K_a}} \right)\]
Thus, \[p{K_a}\] the value decreases with an increase in the acid dissociation constant..
Complete step by step answer:
The ionization of phenol can be written as,
\[{C_6}{H_5}OH \rightleftarrows {C_6}{H_5}{O^ - } + {H^ + }\]
Phenoxide ion is more stable than phenol. So phenol readily loses its proton and hence it is acidic.
It is known that \[{K_a}\] is the acid dissociation constant and it is the equilibrium constant for the dissociation of acid. When the compound easily dissociates and gives up the proton, then it has a high \[{K_a}\] value.
Generally, the acidity of the compound increases with an increase in \[{K_a}\] value. Since the higher the \[{K_a}\]value, the easier will be the dissociation and it will ease the proton removal from the compound. Thus, it will increase its acidity.
It is also known that \[p{K_a}\] is the negative logarithm of the acid dissociation constant.
\[p{K_a} = - \log \left( {{K_a}} \right)\]
Thus, when \[{K_a}\] of the compound is increases, then its \[p{K_a}\] value decreases since it is the negative logarithm of \[{K_a}\]
Therefore, the acidity of the compound increases with an increase in \[{K_a}\] value thereby, decreasing its \[p{K_a}\] value and it decreases with a decrease in \[{K_a}\] thereby, increasing its \[p{K_a}\] value.
Thus, the acidic strength of phenol decreases with an increase in \[p{K_a}\] value
So, the correct answer is Option A.
Note: Similar to acid dissociation, the dissociation constant for the base can be written as \[{K_b}\]. \[p{K_b}\]is the negative logarithm of base dissociation constant. Higher the\[{K_b}\], greater will be the basicity of the compound. The equation which relates \[p{K_a}\] and \[p{K_b}\] can be written as,\[p{K_a} + p{k_b} = 14\]. Thus, if one value of the dissociation constant is known, then another dissociation constant can be calculated.
Whenever the compound tends to readily lose its proton by dissociation, then it indicates that the compound is highly acidic. \[p{K_a}\] is the negative logarithm of the acid dissociation constant.
I.e.\[p{K_a} = - \log \left( {{K_a}} \right)\]
Thus, \[p{K_a}\] the value decreases with an increase in the acid dissociation constant..
Complete step by step answer:
The ionization of phenol can be written as,
\[{C_6}{H_5}OH \rightleftarrows {C_6}{H_5}{O^ - } + {H^ + }\]
Phenoxide ion is more stable than phenol. So phenol readily loses its proton and hence it is acidic.
It is known that \[{K_a}\] is the acid dissociation constant and it is the equilibrium constant for the dissociation of acid. When the compound easily dissociates and gives up the proton, then it has a high \[{K_a}\] value.
Generally, the acidity of the compound increases with an increase in \[{K_a}\] value. Since the higher the \[{K_a}\]value, the easier will be the dissociation and it will ease the proton removal from the compound. Thus, it will increase its acidity.
It is also known that \[p{K_a}\] is the negative logarithm of the acid dissociation constant.
\[p{K_a} = - \log \left( {{K_a}} \right)\]
Thus, when \[{K_a}\] of the compound is increases, then its \[p{K_a}\] value decreases since it is the negative logarithm of \[{K_a}\]
Therefore, the acidity of the compound increases with an increase in \[{K_a}\] value thereby, decreasing its \[p{K_a}\] value and it decreases with a decrease in \[{K_a}\] thereby, increasing its \[p{K_a}\] value.
Thus, the acidic strength of phenol decreases with an increase in \[p{K_a}\] value
So, the correct answer is Option A.
Note: Similar to acid dissociation, the dissociation constant for the base can be written as \[{K_b}\]. \[p{K_b}\]is the negative logarithm of base dissociation constant. Higher the\[{K_b}\], greater will be the basicity of the compound. The equation which relates \[p{K_a}\] and \[p{K_b}\] can be written as,\[p{K_a} + p{k_b} = 14\]. Thus, if one value of the dissociation constant is known, then another dissociation constant can be calculated.