
How do you write the chemical equation for the acid-ionization equilibrium of acetic acid in water?
Answer
550.8k+ views
Hint: To approach this question we should first know how an acid gets dissociated or ionised in the aqueous solution. And we write the acid –ionisation equilibrium expression with the aid of the concentration of the ions formed in the aqueous solution.
Complete step by step answer:
- So the question is asked, how will we write the chemical equation for the acid-ionisation equilibrium of the weak acid acetic acid in aqueous medium.Before writing the equation first we should understand the concept clearly.
- From the lower classes we are classifying substances as base and acid.We are familiar with the fact that those substance which furnish protons or say ${{H}^{+}}$ ions in the aqueous solution and base are those which accepts the ${{H}^{+}}$ ions or protons from the solution.
- So now we know that the acids furnish protons in the solution. And the question is how do they furnish the protons.
- Acids on dissolution with aqueous medium the acids ionises and furnish the protons, in this reaction we can say that the water is acting as the base as it accepts the protons furnished by acid and forms hydronium ion.
- Now let’s move to the solution part of the question, here the acid taken is acetic acid and we all know that acetic acid is a weak acid and it does not completely dissociate in the aqueous medium and it can carry out backward reaction ie it could accept the proton and get back to the undissociated form. So the dissociation of acetic acid is said to be in equilibrium.
So we now know that the ionisation of acetic acid is an equilibrium reaction.Let's write the equation for the ionisation reaction,
$C{{H}_{3}}COOH+{{H}_{2}}O\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}_{3}}{{O}^{+}}$
Now let’s write the acid dissociation constant ie the expression for acid-ionisation equilibrium value which is represent as ,${{K}_{a}}$
${{K}_{a}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}_{3}}{{O}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}$
At standard conditions the acid dissociation constant of acetic acid has a value of ${{K}_{a}}=1.74\times {{10}^{-5}}$
Note: While writing the acid dissociation constant expression we should check whether the mass is balanced on the both sides of the equation. And also check the sum of charges on the reactant side and on the product side is equal or not.
- The concentration of water is not written in the dissociation constant since the activity of water is 1 since it is a pure substance and the concentration cannot be altered.
Complete step by step answer:
- So the question is asked, how will we write the chemical equation for the acid-ionisation equilibrium of the weak acid acetic acid in aqueous medium.Before writing the equation first we should understand the concept clearly.
- From the lower classes we are classifying substances as base and acid.We are familiar with the fact that those substance which furnish protons or say ${{H}^{+}}$ ions in the aqueous solution and base are those which accepts the ${{H}^{+}}$ ions or protons from the solution.
- So now we know that the acids furnish protons in the solution. And the question is how do they furnish the protons.
- Acids on dissolution with aqueous medium the acids ionises and furnish the protons, in this reaction we can say that the water is acting as the base as it accepts the protons furnished by acid and forms hydronium ion.
- Now let’s move to the solution part of the question, here the acid taken is acetic acid and we all know that acetic acid is a weak acid and it does not completely dissociate in the aqueous medium and it can carry out backward reaction ie it could accept the proton and get back to the undissociated form. So the dissociation of acetic acid is said to be in equilibrium.
So we now know that the ionisation of acetic acid is an equilibrium reaction.Let's write the equation for the ionisation reaction,
$C{{H}_{3}}COOH+{{H}_{2}}O\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}_{3}}{{O}^{+}}$
Now let’s write the acid dissociation constant ie the expression for acid-ionisation equilibrium value which is represent as ,${{K}_{a}}$
${{K}_{a}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}_{3}}{{O}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}$
At standard conditions the acid dissociation constant of acetic acid has a value of ${{K}_{a}}=1.74\times {{10}^{-5}}$
Note: While writing the acid dissociation constant expression we should check whether the mass is balanced on the both sides of the equation. And also check the sum of charges on the reactant side and on the product side is equal or not.
- The concentration of water is not written in the dissociation constant since the activity of water is 1 since it is a pure substance and the concentration cannot be altered.
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