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 %

Explanation:

We are given:

Molarity of solution = 0.10 M

Let us assume that $K_a$ of the given acid is $6.7\times 10^{-7}$

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:

$K_a=\frac{[H^+][A^-]}{[HA]}$

We are given:

$K_a=6.7\times 10^{-7}$

Putting values in above equation, we get:

$6.7\times 10^{-7}=\frac{x\times x}{0.1-x}\\\\x^2+(6.7\times 10^{-7})x-6.7\times 10^{-8}=0\\\\x=0.00026,-0.00026$

Neglecting the negative value of 'x'.

To calculate the percent ionization, we use the equation:

$\%\text{ ionization}=\frac{[H^+]_{eq}}{[HA]_i}\times 100$

$[H^+]_{eq}=x=0.00026M$

$[HA]_i=0.1M$

Putting values in above equation, we get:

$\%\text{ ionization}=\frac{0.00026}{0.1}\times 100\\\\\%\text{ ionization}=0.26\%$

Hence, the percent ionization of HA is 0.26 %