4 years ago

Calculate the [OH−] value with [H3O+]=3.0×10−9M.

Chemistry

1 answer:

saw5 [17]4 years ago

4 0

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

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$HF\rightarrow H^+F^-$

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 $8.96\times 10^{-5}$

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