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Calculate the percent dissociation of butanoic acid (C3H2CO2H) in a 1.4 mM aqueous solution of the stuff.

Sagot :

Answer: The percent dissociation of butanoic acid is 9.8%

Explanation:

[tex]C_3H_2CO_2H\rightarrow H^+C_3H_2CO_2^-[/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= 1.4 mM = [tex]1.4\times 10^{-3}[/tex] and [tex]\alpha[/tex] = dissociation constant  

[tex]K_a=1.5\times 10^{-5}[/tex]

Putting in the values we get:

[tex]1.5\times 10^{-5}=\frac{(1.4\times 10^{-3}\times \alpha)^2}{(1.4\times 10^{-3}-1.4\times 10^{-3}\times \alpha)}[/tex]

[tex](\alpha)=0.098=9.8\%[/tex]

Thus percent dissociation of butanoic acid is 9.8%

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