
Which of the following is not a buffer solution ?
Answer
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Hint: A buffer solution is a solution containing a weak acid and its conjugate base. Another buffer solution could be of a base and its conjugate acid. Check the compounds given in the options above to identify which option has a weak acid and its conjugate base or vice versa.
Complete step-by-step answer:
As mentioned in the hint above, a solution containing a weak acid and conjugate base or vice versa is called a buffer solution. The pH of the buffer changes very little which is insignificant when a small quantity of strong acid or base is added.
Buffer solutions are used to keep the pH of the solution nearly constant value. This property of buffers is used in many chemical applications.
Buffer solutions exist in nature as well. For example, the bicarbonate buffer system is used to regulate the pH of blood.
Buffer solutions achieve their property of resisting change in pH due to the presence of an equilibrium between a weak acid and its respective conjugate base or vice versa.
$\text{HA }\to \text{ }{{\text{H}}^{\text{+}}}\text{ + }{{\text{A}}^{-}}$
We will now examine the compounds given in the options and determine which one does not form a buffer system.
In (A), $C{{H}_{3}}COOH+C{{H}_{3}}COONa$ is used which is a buffer as ethanoic acid is a weak organic acid.
In (B), ${{H}_{3}}B{{O}_{3}}+N{{a}_{3}}B{{O}_{3}}$ is used which is a buffer as well because orthoboric acid or boric acid is an extremely weak acid that releases hydrogen ions from water molecule.
In (C), $HCl{{O}_{4}}+NaCl{{O}_{4}}$ is used which cannot form a buffer solution. This is because perchloric acid is a very strong acid and its base is a strong electrolyte as well.
In (D), $N{{H}_{4}}OH+{{(N{{H}_{4}})}_{2}}S{{O}_{4}}$ is used which can form a buffer solution. This is because ammonium hydroxide is a weak base used for detection of group III elements in qualitative analysis of elements.
Therefore, the correct answer is option (C).
Note: Buffer capacity is used to measure the resistance of the solution towards change in pH. It is defined as follows:
$\text{ }\!\!\beta\!\!\text{ = }\dfrac{\text{d}{{\text{C}}_{\text{b}}}}{\text{d(pH)}}\text{ = }-\dfrac{\text{d}{{\text{C}}_{a}}}{\text{d(pH)}}$
Where,
$d{{C}_{b}}$ is the infinitesimal amount of strong base added
$d{{C}_{a}}$ is the infinitesimal amount of strong acid added
Complete step-by-step answer:
As mentioned in the hint above, a solution containing a weak acid and conjugate base or vice versa is called a buffer solution. The pH of the buffer changes very little which is insignificant when a small quantity of strong acid or base is added.
Buffer solutions are used to keep the pH of the solution nearly constant value. This property of buffers is used in many chemical applications.
Buffer solutions exist in nature as well. For example, the bicarbonate buffer system is used to regulate the pH of blood.
Buffer solutions achieve their property of resisting change in pH due to the presence of an equilibrium between a weak acid and its respective conjugate base or vice versa.
$\text{HA }\to \text{ }{{\text{H}}^{\text{+}}}\text{ + }{{\text{A}}^{-}}$
We will now examine the compounds given in the options and determine which one does not form a buffer system.
In (A), $C{{H}_{3}}COOH+C{{H}_{3}}COONa$ is used which is a buffer as ethanoic acid is a weak organic acid.
In (B), ${{H}_{3}}B{{O}_{3}}+N{{a}_{3}}B{{O}_{3}}$ is used which is a buffer as well because orthoboric acid or boric acid is an extremely weak acid that releases hydrogen ions from water molecule.
In (C), $HCl{{O}_{4}}+NaCl{{O}_{4}}$ is used which cannot form a buffer solution. This is because perchloric acid is a very strong acid and its base is a strong electrolyte as well.
In (D), $N{{H}_{4}}OH+{{(N{{H}_{4}})}_{2}}S{{O}_{4}}$ is used which can form a buffer solution. This is because ammonium hydroxide is a weak base used for detection of group III elements in qualitative analysis of elements.
Therefore, the correct answer is option (C).
Note: Buffer capacity is used to measure the resistance of the solution towards change in pH. It is defined as follows:
$\text{ }\!\!\beta\!\!\text{ = }\dfrac{\text{d}{{\text{C}}_{\text{b}}}}{\text{d(pH)}}\text{ = }-\dfrac{\text{d}{{\text{C}}_{a}}}{\text{d(pH)}}$
Where,
$d{{C}_{b}}$ is the infinitesimal amount of strong base added
$d{{C}_{a}}$ is the infinitesimal amount of strong acid added
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