Question

A certain weak acid, HA, has a Ka value of 1.8×10−7. Calculate the percent ionization of HA in a 0.10 M solution. Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer: The percent ionization of HA is 0.13 %

Step-by-step explanation:

We are given:

Molarity of HA solution = 0.10 M

The chemical equation for the ionization of HA follows:

`HA\rightarrow H^++A^-`

Initial: 0.1

At eqllm: (0.1-x) x x

The expression of
`K_a` for above equation follows:

`K_a=([H^+][A^-])/([HA])`

We are given:

`K_a=1.8* 10^(-7)`

Putting values in above equation, we get:

`1.8* 10^(-7)=(x* x)/(0.1-x)\\\\x^2+(1.8* 10^(-7))x-1.8* 10^(-8)=0\\\\x=0.00013,-0.00013`

Neglecting the negative value of 'x' because concentration cannot be negative.

To calculate the percent ionization, we use the equation:

`\%\text{ ionization}=([H^+]_(eq))/([HA]_i)* 100`

`[H^+]_(eq)=x=0.00013M`

`[HA]_i=0.1M`

Putting values in above equation, we get:

`\%\text{ ionization}=(0.00013)/(0.1)* 100\\\\\%\text{ ionization}=0.13\%`

Hence, the percent ionization of HA is 0.13 %