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Sagot :
Answer: [tex][H^+][/tex] of 0.056 M HF solution is [tex]8.96\times 10^{-5}[/tex]
Explanation:
[tex]HF\rightarrow H^+F^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.056 M and [tex]\alpha[/tex] = ?
[tex]K_a=1.45\times 10^{-7}[/tex]
Putting in the values we get:
[tex]1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}[/tex]
[tex](\alpha)=0.0016[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex][H^+]=0.056\times 0.0016=8.96\times 10^{-5}[/tex]
Thus [tex][H^+][/tex] of 0.056 M HF solution is [tex]8.96\times 10^{-5}[/tex]
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