Question

Ka for ascorbic acid is 5*10^-5 .Calculate the [h+] in an aq. Solution in which concentration of asc- is 0.02m

Answer 1

Concept :

There is following relation between equilibrium constant and degree of dissociation

$\alpha or H^{+}  = \sqrt{\frac{K_{a} }{C} }$

Here, C denotes the concentration of given acid

Explanation:

On substituting given values in above expression

We get,

$H^{+} = \sqrt{\frac{5*10^{-5} }{0.02} }  = \sqrt{25*10^{-4} } = 0.05$

Conclusion:

Concentration of Hydrogen ions = 0.05

Answer 2

Answer: Concentration of $[H^+]$ is 0.0003M

Solution:

Given: $[Asc^-]=0.02 M,K_a=5\times 10^{-5}$

              $AscH\rightleftharpoons Asc^-+H^+$

initially      c            0            0

at eq'm  $(c-c\alpha )$      $c\alpha$    $c\alpha$  

$K_a=\frac{[H^+][Asc^-]}{[AscH]}$

$K_a=\frac{[c\alpha ][c\alpha ]}{[c-c\alpha ]}$

$K_a=\frac{c{\alpha }^2}{(1-\alpha )}$

$5\times 10^{-5}=\frac{(0.02)(\alpha )^2}{1-\alpha }$

On solving  above equation we get  $\alpha$

$\alpha=0.015$

$[H^+]=c\alpha =0.02\times 0.015=0.0003 M$