Answer
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Hint: We have studied that an aqueous solution of ammonium acetate acts as a buffer solution because its salt is a product of a weak acid (acetic acid ) and a weak base ( ammonium hydroxide). A buffer solution does not show significant change in $pH$ value. when a small amount acid or base is added to it or we can say that the acidic and basic nature of buffer solution is reserved. We can solve this problem by calculating the $pH$ value of ammonium acetate solution which will eventually give information whether the solution is acid, basic, neutral or slightly acidic.
Complete step by step solution:
We know that in ammonium acetate both components are equally hydrolysed so they both counter equally to each other. We know that the value of the dissociation constant of water $({K_w})$ is $1.0 \times {10^{ - 14}}$. We can calculate the dissociation constant of following reactions,
$C{H_3}CO{O^ - } + {H_2}O \rightleftharpoons C{H_3}COOH + O{H^ - }$ $ \Rightarrow K' = \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}$
$N{H^ + }_4 + {H_2}O \rightleftharpoons N{H_4}OH + {H^ + }$ $ \Rightarrow K'' = \dfrac{{{K_w}}}{{{K_b}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}$
Now we can see that $K' = K''$ Hence it results in $[{H^ + }] = [O{H^ - }]$. We will calculate values of $p{K_a},p{K_b}$ which is $p{K_a} = - \log {K_a}$ and $p{K_b} = - \log {K_b}$.
The $pH$ of the salt of a weak acid and weak base can be calculated as; $ \Rightarrow pH = 7 - \dfrac{1}{2}p{K_b} + \dfrac{1}{2}p{K_a} = 7 - \dfrac{{4.74}}{2} + \dfrac{{4.74}}{2} = 7$, when the concentration of ${H^ + }$ and $O{H^ - }$ are the same in the solution so the solution is neutral ,thus the $pH$ is $7$. So option C is correct as the solution is neutral.
Note: The dissociation constant of acid and base is given in the problem. The value of both constants are equals which further gives the equal acidic and basic concentration in the solution and makes it neutral. We have also calculated $pH$ value of salt and it also comes neutral.
Complete step by step solution:
We know that in ammonium acetate both components are equally hydrolysed so they both counter equally to each other. We know that the value of the dissociation constant of water $({K_w})$ is $1.0 \times {10^{ - 14}}$. We can calculate the dissociation constant of following reactions,
$C{H_3}CO{O^ - } + {H_2}O \rightleftharpoons C{H_3}COOH + O{H^ - }$ $ \Rightarrow K' = \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}$
$N{H^ + }_4 + {H_2}O \rightleftharpoons N{H_4}OH + {H^ + }$ $ \Rightarrow K'' = \dfrac{{{K_w}}}{{{K_b}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}$
Now we can see that $K' = K''$ Hence it results in $[{H^ + }] = [O{H^ - }]$. We will calculate values of $p{K_a},p{K_b}$ which is $p{K_a} = - \log {K_a}$ and $p{K_b} = - \log {K_b}$.
The $pH$ of the salt of a weak acid and weak base can be calculated as; $ \Rightarrow pH = 7 - \dfrac{1}{2}p{K_b} + \dfrac{1}{2}p{K_a} = 7 - \dfrac{{4.74}}{2} + \dfrac{{4.74}}{2} = 7$, when the concentration of ${H^ + }$ and $O{H^ - }$ are the same in the solution so the solution is neutral ,thus the $pH$ is $7$. So option C is correct as the solution is neutral.
Note: The dissociation constant of acid and base is given in the problem. The value of both constants are equals which further gives the equal acidic and basic concentration in the solution and makes it neutral. We have also calculated $pH$ value of salt and it also comes neutral.
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