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%

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:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

a) Given c= 0.32M and $K_a=6.3\times 10^{-5}$

$\alpha$ = ?

Putting in the values we get:

$6.3\times 10^{-5}=\frac{(0.32\times \alpha)^2}{(0.32-0.32\times \alpha)}$

$(\alpha)=0.014$

$\%(\alpha)=0.014\times 100=1.4\%$

b) Given c= 0.00014 M and $K_a=6.3\times 10^{-5}$

$\alpha$ = ?

Putting in the values we get:

$6.3\times 10^{-5}=\frac{(0.00014\times \alpha)^2}{(0.00014-0.00014\times \alpha)}$

$(\alpha)=0.48$

$\%(\alpha)=0.48\times 100=48\%$