Question

Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7

Answer 1

Answer:  $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

Explanation:

$HF\rightarrow H^+F^-$

 cM              0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

So dissociation constant will be:

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

Give c= 0.056 M and $\alpha$ = ?

$K_a=1.45\times 10^{-7}$

Putting in the values we get:

$1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}$

$(\alpha)=0.0016$

$[H^+]=c\times \alpha$

$[H^+]=0.056\times 0.0016=8.96\times 10^{-5}$  

Thus $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$