Question

Calculate the volume of N/5 NaOH.
required to neutralize CH₂COOH
which is resulting from hydrolysis
of 4. ug ethyl acetate. C CH₃COOCH = 28​

Answer 1

Answer:

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Answer 2

Answer:

The volume of NaOH is required to neutralize acetic acid will be  is equal to 250ml.

Explanation:

Consider that the solution of $CH_{3}COOH = 1000ml$

The mass of  $CH_{3}COOC_{2}H_{5} =4.4g$

The degree of dissociation $\alpha =1$

Moles of $CH_{3}COOC_{2}H_{5}=\frac{4.4}{88}=0.05 mol$

               $CH_{3}COOC_{2}H_{5}$    ⇄   $CH_{3}COOH+C_{2}H_{5}OH$

Initial  :        $0.05M$                        ----                 ----

Equ. :    $0.05-0.05\alpha$                   $0.05\alpha$              $0.05\alpha$

Final  :           -------                      $0.05M$            $0.05M$

Normality = molarity × n-factor

Normality of  CH₃COOH$= 0.05 *1=0.05N$              [∵ n-factor=1]

We know that

$N_{NaOH}*V_{NaOH}=N_{CH_{3}COOH}*V_{CH_{3}COOH}$

$(\frac{1}{5} )(V_{NaOH})=(0.05)(1000)\\V_{NaOH}=(0.05)(1000)(5)\\V_{NaOH}=250ml$

Therefore  volume of NaOH will be added to neutalization, is equal to  250ml.