At $30^\circ$C , the degree of dissociation of 0.066M HA is 0.0145. What would be the degree of dissociation of 0.02M solution of the acid at the same temperature?
Answer
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Hint: The relation between degree of dissociation and dissociation constant is given by \[\alpha = \sqrt {({K_a}/C)} \].
The value of dissociation constant of a particular acid is constant at constant temperature.
Complete answer:
Degree of dissociation depends on electrolyte concentration which decreases with increase in concentration. The relation between dissociation constant and degree of dissociation is given by,
\[K = \dfrac{{{\alpha ^2}C}}{{1 - \alpha }}\]
For weak acid, the degree of dissociation is very small, so that the degree of dissociation is related to the dissociation constant by the following formula.
\[\alpha = \sqrt {({K_a}/C)} \]
\[ \Rightarrow {K_a} = {\alpha ^2}C\]
Where, \[\alpha \]\[ = \] degree of dissociation
\[{K_a}\]\[ = \] dissociation constant
\[C = \] concentration of acid
In the question, it is given that the degree of dissociation \[ = 0.0145\]
Concentration of the acid \[ = 0.066M\]
Substituting the values in the above formula, the dissociation constant will be,
\[ \Rightarrow {K_a} = {\alpha ^2}C = {(0.0145)^2} \times 0.066 = 1.39 \times {10^{ - 5}}M\]
The temperature of the system remains the same, so, the dissociation constant does not change at all.
Now, the concentration of the acid \[ = 0.02M\]
So, the degree of dissociation will be,
\[\alpha = \sqrt {({K_a}/C)} = {(\dfrac{{1.39 \times {{10}^{ - 5}}}}{{0.02}})^{1/2}} = 2.63 \times {10^{ - 2}}\] = 0.0263
Hence, the degree of dissociation will be 0.0263.
Note: Compounds that completely dissociate are called strong electrolyte having high degree of dissociation, whereas weak electrolytes have negligible degree of dissociation. The above relation between degree of dissociation and dissociation constant is known as Ostwald’s dilution law.
For exothermic dissolution, the dissociation constant increases with increase in temperature.
For endothermic dissolution, the dissociation constant decreases with increase in temperature.
The value of dissociation constant of a particular acid is constant at constant temperature.
Complete answer:
Degree of dissociation depends on electrolyte concentration which decreases with increase in concentration. The relation between dissociation constant and degree of dissociation is given by,
\[K = \dfrac{{{\alpha ^2}C}}{{1 - \alpha }}\]
For weak acid, the degree of dissociation is very small, so that the degree of dissociation is related to the dissociation constant by the following formula.
\[\alpha = \sqrt {({K_a}/C)} \]
\[ \Rightarrow {K_a} = {\alpha ^2}C\]
Where, \[\alpha \]\[ = \] degree of dissociation
\[{K_a}\]\[ = \] dissociation constant
\[C = \] concentration of acid
In the question, it is given that the degree of dissociation \[ = 0.0145\]
Concentration of the acid \[ = 0.066M\]
Substituting the values in the above formula, the dissociation constant will be,
\[ \Rightarrow {K_a} = {\alpha ^2}C = {(0.0145)^2} \times 0.066 = 1.39 \times {10^{ - 5}}M\]
The temperature of the system remains the same, so, the dissociation constant does not change at all.
Now, the concentration of the acid \[ = 0.02M\]
So, the degree of dissociation will be,
\[\alpha = \sqrt {({K_a}/C)} = {(\dfrac{{1.39 \times {{10}^{ - 5}}}}{{0.02}})^{1/2}} = 2.63 \times {10^{ - 2}}\] = 0.0263
Hence, the degree of dissociation will be 0.0263.
Note: Compounds that completely dissociate are called strong electrolyte having high degree of dissociation, whereas weak electrolytes have negligible degree of dissociation. The above relation between degree of dissociation and dissociation constant is known as Ostwald’s dilution law.
For exothermic dissolution, the dissociation constant increases with increase in temperature.
For endothermic dissolution, the dissociation constant decreases with increase in temperature.
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