Question

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.

Answer 1

Answer:
`K_a` for the acid is
`3.34* 10^(-4)`

Step-by-step explanation:

`HC_9H_7O_4\rightarrow H^+C_9H_7O_4^-`

cM 0 0

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

Give c = 0.328 M and
`pH=1.987`

`1.987=-log[H^+]`

`[H^+]=0.0103`

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

`0.0103=0.328* \alpha`

`\alpha=0.0314`

So dissociation constant will be:

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

Putting in the values we get:

`K_a=((0.328* 0.0314)^2)/((0.328-0.328* 0.0314))`

`K_a=3.34* 10^(-4)`