Question

Wul2 naviya What is the percent dissociation of 0.10 M phenol? (The Ka value for phenolis 1.3 x 10-0.) ution 5 Complete out of Answer: tion Check Finish attemp

Answer 1

Let's see percent dissociation of phenol =

Ka=1.3x 10 ^-10

C₆H₅OH + H₂O   $\LARGE \leftrightarrow$ C₆H₅ O^- + H₃O⁺

Remarks	conc. (C6H5OH)	conc.C6H5O-	conc . H3O+
Initial	0.10 M	0	0
Conc.	-X	+X	+X
Equilibrium	0.10M -X	X	X

Based on above equation ,

Ka= (C₆H₅O^-) x (H₃O⁺) / (C₆H₅OH)

= ( x²)/ (0.10 -x)

We know that Ka= 1.3 x 10^-10

by putting values ,

So , (x²) / (0.10 -x) = 1.3 x 10 ^-10

# But (0.10 -x )= very small ,so value of' x' can be neglected .

(x²) = (0.10 x 1.3 x 10^-10)

(x²) = 0.13 x 10^-12

( X) = 0.3605 x 10^-6 M

But here we have to calculate percent dissociation =

= (amount dissociated ( M)/ (initial concentration( M) x 100

= (0.3 x 10^-6 M ) / (0.1M) x100

=0.003 %