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In the laboratory, a general chemistry student measured the pH of a 0.328 M aqueous solution of acetylsalicylic acid (aspirin), HC9H7O4 to be 1.987. Use the information she obtained to determine the Ka for this acid.

Sagot :

Answer: [tex]K_a[/tex] for the acid is [tex]3.34\times 10^{-4}[/tex]

Explanation:

[tex]HC_9H_7O_4\rightarrow H^+C_9H_7O_4^-[/tex]

 cM              0             0

[tex]c-c\alpha[/tex]        [tex]c\alpha[/tex]          [tex]c\alpha[/tex]  

Give c = 0.328 M and [tex]pH=1.987[/tex]

[tex]1.987=-log[H^+][/tex]

[tex][H^+]=0.0103[/tex]

[tex][H^+]=c\times \alpha[/tex]

[tex]0.0103=0.328\times \alpha[/tex]

[tex]\alpha=0.0314[/tex]

So dissociation constant will be:

[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Putting in the values we get:

[tex]K_a=\frac{(0.328\times 0.0314)^2}{(0.328-0.328\times 0.0314)}[/tex]

[tex]K_a=3.34\times 10^{-4}[/tex]

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