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GREYUIT [131]
3 years ago
6

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 a

nswer to two significant figures and include the appropriate units.
Chemistry
1 answer:
Marta_Voda [28]3 years ago
8 0

<u>Answer:</u> The percent ionization of HA is 0.13 %

<u>Explanation:</u>

We are given:

Molarity of HA solution = 0.10 M

The chemical equation for the ionization of HA follows:

                     HA\rightarrow H^++A^-

<u>Initial:</u>            0.1

<u>At eqllm:</u>      (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=1.8\times 10^{-7}

Putting values in above equation, we get:

1.8\times 10^{-7}=\frac{x\times x}{0.1-x}\\\\x^2+(1.8\times 10^{-7})x-1.8\times 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}=\frac{[H^+]_{eq}}{[HA]_i}\times 100

[H^+]_{eq}=x=0.00013M

[HA]_i=0.1M

Putting values in above equation, we get:

\%\text{ ionization}=\frac{0.00013}{0.1}\times 100\\\\\%\text{ ionization}=0.13\%

Hence, the percent ionization of HA is 0.13 %

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