Respuesta :
Answer: The percent ionization of HA is 0.26 %
Explanation:
We are given:
Molarity of solution = 0.10 M
Let us assume that [tex]K_a[/tex] of the given acid is [tex]6.7\times 10^{-7}[/tex]
The chemical equation for the ionization of HA follows:
[tex]HA\rightarrow H^++A^-[/tex]
Initial: 0.1
At eqllm: (0.1-x) x x
The expression of [tex]K_a[/tex] for above equation follows:
[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]
We are given:
[tex]K_a=6.7\times 10^{-7}[/tex]
Putting values in above equation, we get:
[tex]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[/tex]
Neglecting the negative value of 'x'.
To calculate the percent ionization, we use the equation:
[tex]\%\text{ ionization}=\frac{[H^+]_{eq}}{[HA]_i}\times 100[/tex]
[tex][H^+]_{eq}=x=0.00026M[/tex]
[tex][HA]_i=0.1M[/tex]
Putting values in above equation, we get:
[tex]\%\text{ ionization}=\frac{0.00026}{0.1}\times 100\\\\\%\text{ ionization}=0.26\%[/tex]
Hence, the percent ionization of HA is 0.26 %
