
A $100ml$ , $0.1M$ solution of ammonium acetate is diluted by adding $100ml$ of water. The $pH$ of the resulting solution will be ( $p{K_a}$ of acetic acid is nearly equal to $p{K_b}$ of $N{H_4}OH$ ):
A. $4.9$
B. $5.0$
C. $7.0$
D. $10.0$
Answer
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Hint: Ammonium acetate is a salt which is prepared by using weak base and weak acid such as acetic acid and ammonium hydroxide. And when such salt is diluted with water then we use the relation given below to determine the $pH$ of the resulting solution.
Formula used:
$pH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_b}$
Where, $pH$ = potential of hydrogen of the solution
$p{K_a}$ = Strength of acid in the solution
$p{K_b}$ = Strength of base in the solution
Complete step by step answer:
Ammonium acetate is salt made up of acetic acid and ammonium hydroxide. Acetic acid is a weak acid and also ammonium hydroxide is a weak base.
The reaction takes place as:
$C{H_3}COOH + N{H_3}OH \to N{H_4}C{H_3}C{O_2} + {H_2}O$
When this solution is diluted with $100ml$ of water then the $pH$ of solution is given by:
$pH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_b}$
Where, $pH$ = potential of hydrogen of the solution
$p{K_a}$ = Strength of acid in the solution
$p{K_b}$ = Strength of base in the solution
According to the question,
The strength of acetic acid is nearly equal to the strength of ammonium hydroxide.
So, we can write:
$ \Rightarrow p{K_a} \simeq p{K_b}$
On substituting the values in the formula we will get:
$pH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_a}$
On further solving it we will get:
$\
\Rightarrow pH = 7 + 0 \\
\Rightarrow pH = 7 \\
\ $
Hence, option C is correct.
Note:
At room temperature${25^0}C$ , solutions which have $pH$ less than $7$ are acidic in nature and solutions which have $pH$ greater than $7$ are basic in nature. Solutions which have $pH$ exactly equal to $7$ at room temperature are neutral in nature for example water. The neutral value of the $pH$ depends on the temperature, being lower than $7$ if the temperature increases. The $pH$ value can be less than $0$ for very strong acids, or greater than $14$ for very strong bases.
Formula used:
$pH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_b}$
Where, $pH$ = potential of hydrogen of the solution
$p{K_a}$ = Strength of acid in the solution
$p{K_b}$ = Strength of base in the solution
Complete step by step answer:
Ammonium acetate is salt made up of acetic acid and ammonium hydroxide. Acetic acid is a weak acid and also ammonium hydroxide is a weak base.
The reaction takes place as:
$C{H_3}COOH + N{H_3}OH \to N{H_4}C{H_3}C{O_2} + {H_2}O$
When this solution is diluted with $100ml$ of water then the $pH$ of solution is given by:
$pH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_b}$
Where, $pH$ = potential of hydrogen of the solution
$p{K_a}$ = Strength of acid in the solution
$p{K_b}$ = Strength of base in the solution
According to the question,
The strength of acetic acid is nearly equal to the strength of ammonium hydroxide.
So, we can write:
$ \Rightarrow p{K_a} \simeq p{K_b}$
On substituting the values in the formula we will get:
$pH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_a}$
On further solving it we will get:
$\
\Rightarrow pH = 7 + 0 \\
\Rightarrow pH = 7 \\
\ $
Hence, option C is correct.
Note:
At room temperature${25^0}C$ , solutions which have $pH$ less than $7$ are acidic in nature and solutions which have $pH$ greater than $7$ are basic in nature. Solutions which have $pH$ exactly equal to $7$ at room temperature are neutral in nature for example water. The neutral value of the $pH$ depends on the temperature, being lower than $7$ if the temperature increases. The $pH$ value can be less than $0$ for very strong acids, or greater than $14$ for very strong bases.
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