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* 10^(-5)`

Step-by-step explanation:

`HF\rightarrow H^+F^-`

cM 0 0

`c-c\alpha`
`c\alpha`
`c\alpha`

So dissociation constant will be:

`K_a=((c\alpha)^(2))/(c-c\alpha)`

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

`K_a=1.45* 10^(-7)`

Putting in the values we get:

`1.45* 10^(-7)=((0.056* \alpha)^2)/((0.056-0.056* \alpha))`

`(\alpha)=0.0016`

`[H^+]=c* \alpha`

`[H^+]=0.056* 0.0016=8.96* 10^(-5)`

Thus
`[H^+]` of 0.056 M HF solution is
`8.96* 10^(-5)`