
A weak acid reacts with a strong base; ionisation constant of a weak acid is \[{10^{ - 4}}\]. Find out equilibrium constant for this reaction.
A.${10^{ - 10}}$
B.${10^{10}}$
C.${10^{ - 9}}$
D.${10^9}$
Answer
570.3k+ views
Hint: The equilibrium constant is used to determine the relationship between products and reactants of a reaction at equilibrium concerning a specific unit.
Complete step by step answer:
Equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. The dissociation of a weak acid can be written as:
\[HA \rightleftharpoons {H^ + } + {A^ - }\].
$\therefore $ The ionic product from the definition of equilibrium constant, \[{{\text{K}}_{\text{a}}} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}\]
We are given the value of \[{{\text{K}}_{\text{a}}} = {10^{ - 4}}\].
Dissociation of water can be written as: \[{H_2}O \rightleftharpoons O{H^ - } + {H^ + }\].
As water is neutral the ionic product for dissociation of water can be written as \[{{\text{K}}_{\text{w}}} = [{H^ + }][O{H^ - }] = {10^{ - 14}}\].
For the reaction of strong Base \[{\text{BOH}}\]with weak acid reaction can be written as:
\[HA + O{H^ - } \rightleftharpoons {H_2}O + {A^ - }\].
Now we can write the dissociation constant as:
\[{\text{K}} = \dfrac{{[{A^ - }][{H_2}O]}}{{[O{H^ - }][HA]}}\]. Finally substituting the value of \[{{\text{K}}_{\text{w}}} = {10^{ - 14}}\]and\[{{\text{K}}_{\text{a}}} = {10^{ - 4}}\] we have:
\[{\text{K = }}\dfrac{{{{\text{K}}_{\text{a}}}}}{{{{\text{K}}_{\text{w}}}}}{\text{ }} = \dfrac{{{{10}^{ - 4}}}}{{{{10}^{ - 14}}}} = {10^{10}}\]
Therefore, we can conclude that the correct answer to this question is option B.
Note:
The difference between strong bases and weak bases: a strong base is a base that is $100\% $ ionized in solution but if it is less than $100\% $ ionized in solution, it is a weak base. There are very few strong bases and certain salts will also affect the acidity or basicity of aqueous solutions because some of the ions will undergo hydrolysis. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyse, while salts with ions that are part of weak acids or bases will hydrolyse.
Complete step by step answer:
Equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. The dissociation of a weak acid can be written as:
\[HA \rightleftharpoons {H^ + } + {A^ - }\].
$\therefore $ The ionic product from the definition of equilibrium constant, \[{{\text{K}}_{\text{a}}} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}\]
We are given the value of \[{{\text{K}}_{\text{a}}} = {10^{ - 4}}\].
Dissociation of water can be written as: \[{H_2}O \rightleftharpoons O{H^ - } + {H^ + }\].
As water is neutral the ionic product for dissociation of water can be written as \[{{\text{K}}_{\text{w}}} = [{H^ + }][O{H^ - }] = {10^{ - 14}}\].
For the reaction of strong Base \[{\text{BOH}}\]with weak acid reaction can be written as:
\[HA + O{H^ - } \rightleftharpoons {H_2}O + {A^ - }\].
Now we can write the dissociation constant as:
\[{\text{K}} = \dfrac{{[{A^ - }][{H_2}O]}}{{[O{H^ - }][HA]}}\]. Finally substituting the value of \[{{\text{K}}_{\text{w}}} = {10^{ - 14}}\]and\[{{\text{K}}_{\text{a}}} = {10^{ - 4}}\] we have:
\[{\text{K = }}\dfrac{{{{\text{K}}_{\text{a}}}}}{{{{\text{K}}_{\text{w}}}}}{\text{ }} = \dfrac{{{{10}^{ - 4}}}}{{{{10}^{ - 14}}}} = {10^{10}}\]
Therefore, we can conclude that the correct answer to this question is option B.
Note:
The difference between strong bases and weak bases: a strong base is a base that is $100\% $ ionized in solution but if it is less than $100\% $ ionized in solution, it is a weak base. There are very few strong bases and certain salts will also affect the acidity or basicity of aqueous solutions because some of the ions will undergo hydrolysis. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyse, while salts with ions that are part of weak acids or bases will hydrolyse.
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