Question

The ph of a 0.01 m solution of a weak acid having degree of dissociation 12.5% is

Answer 1

Answer : The pH will be, 2.904

Solution :  Given,

Concentration (C) = 0.01 M

Degree of dissociation $(\alpha)$ = 12.5% = 0.125

The equilibrium reaction for dissociation of weak acid is,

                           $HA\rightleftharpoons A^-+H^+$

initially conc.         c       0      0

At eqm.           $c(1-\alpha)$   $c\alpha$   $c\alpha$

First we have to calculate the concentration of $[H^+]$

$[H^+]=c\alpha$

$[H^+]=(0.01)\times (0.125)=1.25\times 10^{-3}M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.25\times 10^{-3})$

$pH=2.904$

Therefore, the pH will be, 2.904

Answer 2

Answer : The pH will be, 2.904

Solution :  Given,

Concentration (C) = 0.01 M

Degree of dissociation  = 12.5% = 0.125

The equilibrium reaction for dissociation of weak acid is,

                         

initially conc.         c       0      0

At eqm.                

First we have to calculate the concentration of

Now we have to calculate the pH.

Therefore, the pH will be, 2.904