Question

(b) Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq) ? Justify your answer including the calculation of percent ionization for each solution.

Answer 1

The 0.10M solution of HC2H3O2(aq) has a higher percent ionization of the acid than the 0.010M solution.

Step-by-step explanation:

In order to determine which solution has a higher percent ionization of the acid, we need to calculate the percent ionization for each solution.

The percent ionization of an acid is the ratio of the concentration of the ionized acid to the initial concentration of the acid, multiplied by 100.

For the 0.10M solution of HC2H3O2(aq), if it is 3% ionized, then the percent ionization would be (3/100) x 100 = 3%.

For the 0.010M solution of HC2H3O2(aq), if it is 1% ionized, then the percent ionization would be (1/100) x 100 = 1%.

Therefore, the 0.10M solution of HC2H3O2(aq) has a higher percent ionization of the acid than the 0.010M solution.

Answer 2

Answer: The percent ionization is higher for 0.010 M solution of
`HC_2H_3O_2`

Step-by-step explanation:

`HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-`

cM 0M 0M

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

So dissociation constant will be:

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

a) Given c= 0.10 M and
`\alpha` = ?

`K_a=1.8* 10^(-5)`

Putting in the values we get:

`1.8* 10^(-5)=((0.10* \alpha)^2)/((0.10-0.10* \alpha))`

`(\alpha)=0.013`

`\%(\alpha)=0.013* 100=1.3\%`

b) Given c= 0.010 M and
`\alpha` = ?

`K_a=1.8* 10^(-5)`

Putting in the values we get:

`1.8* 10^(-5)=((0.010* \alpha)^2)/((0.010-0.010* \alpha))`

`(\alpha)=0.041`

`\%(\alpha)=0.041* 100=4.1\%`

The percent ionization is higher for 0.010 M solution of
`HC_2H_3O_2`