Question

Calculate the percent ionization of benzoic acid at the following concentrations. (a) 0.32 M WebAssign will check your answer for the correct number of significant figures. 1.4 Correct: Your answer is correct. (b) 0.00014 M WebAssign will check your answer for the correct number of significant figures. 48 Correct: Your answer is correct.

Answer 1

Answer: a) 1.4 %

b) 48%

Step-by-step explanation:

`C_7H_6O_2\rightarrow H^+C_7H_5O_2^-`

cM 0 0

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

So dissociation constant will be:

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

a) Given c= 0.32M and
`K_a=6.3* 10^(-5)`

`\alpha` = ?

Putting in the values we get:

`6.3* 10^(-5)=((0.32* \alpha)^2)/((0.32-0.32* \alpha))`

`(\alpha)=0.014`

`\%(\alpha)=0.014* 100=1.4\%`

b) Given c= 0.00014 M and
`K_a=6.3* 10^(-5)`

`\alpha` = ?

Putting in the values we get:

`6.3* 10^(-5)=((0.00014* \alpha)^2)/((0.00014-0.00014* \alpha))`

`(\alpha)=0.48`

`\%(\alpha)=0.48* 100=48\%`