Question

0.1 mol each of ethyl alcohol and acetic acid one allowed to react and at equilibrium the acid was exactly neutralized by 100ml of 0.75 n naoh

Answer 1

Answer:

Answer

{ CH }_{ 3 }COOH+{ C }_{ 2 }{ H }_{ 5 }OH\rightleftharpoons { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }OCH

3

COOH+C

2

H

5

OH⇌CH

3

COOC

2

H

5

+H

2

O

0.1 0.1 0 0 Before reaction

(0.1-x)(0.1−x) (0.1-x)(0.1−x) xx xx

mEq of of acetic aid left = mEq of NaOH used = 100 x 0.85 = 85

\therefore∴ Millimoles of aetic acid left = 85 (\because \quad mono\quad basic)(∵monobasic)

\therefore Moles\, of \,acetic\, acid\, left = 0.085∴ Molesofaceticacidleft=0.085

(0.01-x)=0.085(0.01−x)=0.085

x=0.015x=0.015

Now,

$${ K }_{ C }=\displaystyle\frac { { x }^{ 2 } }{ { (0.1-x) }^{ 2 } }

=\frac { { (0.015) }^{ 2 } }{ { (0.085) }^{ 2 } } =0.031$$

Answer 2

Answer:

What's your question dear?? Y did u tell us this??