Question

In a 0.100 m hf solution, the percent dissociation is determined to be 9.5%. calculate the ka for hf based on this data.

Answer 1

Answer : The dissociation constant (Ka) = 9.025 × $10^{-4}$

Solution :  Given,

                Concentration HF solution = 0.100 M

                 % Dissociation = 9.5 %

The equation for dissociation of HF is :

                  $HF \rightleftharpoons H^{+}+ F^{-}$

The Ka expression for HF is :

$Ka=\frac{[H^{+}][F^{-}]}{[HF]}$        ............. (1)

Step 1 : we find the $[H^{+}]$ by using the concentration and % dissociation.

 $[H^{+}]$ = Concentration HF solution ×  % Dissociation

 $[H^{+}]$ = 0.100 M × $\frac{9.5}{100}$ = 9.5 × $10^{-3}$ M

Step 2 : For $[F^{-}]$ , the concentration of  $[F^{-}]$ is equal to the  $[H^{+}]$. From the above equation the stoichiometry of   $[F^{-}]$ and $[H^{+}]$ is 1:1.

Therefore,

$[F^{-}]$ =  $[H^{+}]$  = 0.100 M × $\frac{9.5}{100}$ =    =  9.5 × $10^{-3}$ M

         

Now, put all the values in equation (1), we get

$Ka=\frac{(9.5\times10^{-3})\times(9.5\times10^{-3})}{(0.1)}$

           = 9.025 × $10^{-4}$