
In the hydrolytic equilibrium A$^{-}$ + H$_2$O $\rightleftharpoons$ HA + OH$^{-}$, K$_a$ = 1.0 $\times$ 10 $^{-5}$ . The degree of hydrolysis of 0.001 M solution of the salt is:
a.) 10 $^{-3}$
b.) 10 $^{-4}$
c.) 10 $^{-5}$
d.) 10 $^{-6}$
Answer
585.9k+ views
Hint: The general reaction of hydrolysis, and the value of dissociation constant is given in the question. The value of dissociation constant for water is 10 $^{-14}$. We can calculate the hydrolysis constant, and the degree of hydrolysis would be calculated further.
Complete step-by-step answer:
First, we will define the ionic product of water. In this we consider the multiplication of the concentration of hydrogen ions, and hydroxide ions at particular temperature.
Now, as mentioned the given reaction is the general reaction of hydrolysis of a salt, and here HA is a weak acid.
We know that the hydrolysis constant, concentration of water, or dissociation constant are interrelated to each other
Thus, the relation can be expressed as:
K$_h$ = $\dfrac{K_w}{K_a}$
Here, K$_h$ represents the hydrolysis constant, K$_w$ represents the concentration of water, or dissociation constant of water, and K$_a$ represents the dissociation constant of the given equation.
Now, we have the value of K$_w$, and K$_a$, i.e. 10 $^{-14}$ and 1.0 $\times$ 10 $^{-5}$ respectively.
If we substitute the respective values to calculate the hydrolysis constant, then
K$_h$ = $\dfrac{10^{-14}}{1 \times 10 ^{-5}}$ = 10 $^{-9}$ .
Thus, the hydrolysis constant is 10 $^{-9}$.
Now, we will calculate the value of the degree of hydrolysis. The expression is given by: h = $\sqrt{\dfrac{K_h}{C}}$
Here, h represents the degree of hydrolysis, and C represents the concentration of solution given in the question i.e. 0.001 M
Let us substitute the values, then we get
h = $\sqrt{\dfrac{10^{-9}}{0.001}}$
h =10 $^{-3}$.
In the last, we can conclude that the degree of hydrolysis of 0.001 M solution of the salt is10 $^{-3}$.
Hence, the correct option is (A).
Note:
Don’t get confused in the expressions related to the degree of hydrolysis. Solve the question step by step. We should remember the expressions given for the general reactions, so it will be easier to solve the question.
Complete step-by-step answer:
First, we will define the ionic product of water. In this we consider the multiplication of the concentration of hydrogen ions, and hydroxide ions at particular temperature.
Now, as mentioned the given reaction is the general reaction of hydrolysis of a salt, and here HA is a weak acid.
We know that the hydrolysis constant, concentration of water, or dissociation constant are interrelated to each other
Thus, the relation can be expressed as:
K$_h$ = $\dfrac{K_w}{K_a}$
Here, K$_h$ represents the hydrolysis constant, K$_w$ represents the concentration of water, or dissociation constant of water, and K$_a$ represents the dissociation constant of the given equation.
Now, we have the value of K$_w$, and K$_a$, i.e. 10 $^{-14}$ and 1.0 $\times$ 10 $^{-5}$ respectively.
If we substitute the respective values to calculate the hydrolysis constant, then
K$_h$ = $\dfrac{10^{-14}}{1 \times 10 ^{-5}}$ = 10 $^{-9}$ .
Thus, the hydrolysis constant is 10 $^{-9}$.
Now, we will calculate the value of the degree of hydrolysis. The expression is given by: h = $\sqrt{\dfrac{K_h}{C}}$
Here, h represents the degree of hydrolysis, and C represents the concentration of solution given in the question i.e. 0.001 M
Let us substitute the values, then we get
h = $\sqrt{\dfrac{10^{-9}}{0.001}}$
h =10 $^{-3}$.
In the last, we can conclude that the degree of hydrolysis of 0.001 M solution of the salt is10 $^{-3}$.
Hence, the correct option is (A).
Note:
Don’t get confused in the expressions related to the degree of hydrolysis. Solve the question step by step. We should remember the expressions given for the general reactions, so it will be easier to solve the question.
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