Question

Calculate the volume of N/5 NaOH required to neutralize CH3COOOH which is resulting from the hydrolysis of 4.4g ethyl acetate (CH3COOC2H5).​

Answer 1

Answer:

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Explanation:

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Answer 2

250ml

250 millilitres of NaOH are needed to neutralise one gramme of acetic acid.

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  :     0.05 M                              --                   --

Equal :       $0.05 -0.05\alpha$                    $0.05\alpha$               $0.05\alpha$

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

Normality = molarity × n-factor

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

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

$\frac{1}{5} V_{NaOH} = 0.05$ × $1000$

$V_{NaOH} = 0.05$ ×$1000$×$5$

$V_{NaOH} = 250 ml$

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