Question

Calculate the percent ionization of ha in a 0.10 m solution. express your answer as a percent using two significant figures.

Answer 1

the percent would be 0.31% hope that helps

Answer 2

Answer: The percent ionization of HA is 0.26 %

Step-by-step explanation:

We are given:

Molarity of solution = 0.10 M

Let us assume that
`K_a` of the given acid is
`6.7* 10^(-7)`

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=6.7* 10^(-7)`

Putting values in above equation, we get:

`6.7* 10^(-7)=(x* x)/(0.1-x)\\\\x^2+(6.7* 10^(-7))x-6.7* 10^(-8)=0\\\\x=0.00026,-0.00026`

Neglecting the negative value of 'x'.

To calculate the percent ionization, we use the equation:

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

`[H^+]_(eq)=x=0.00026M`

`[HA]_i=0.1M`

Putting values in above equation, we get:

`\%\text{ ionization}=(0.00026)/(0.1)* 100\\\\\%\text{ ionization}=0.26\%`

Hence, the percent ionization of HA is 0.26 %