When the 0.10 m solution of hclo2 is allowed to come to equilibrium, what percent of the acid molecules are ionized?

1 answer:

4 0

If 0.10 m solution of hclo2 is allowed to come to equilibrium, the percent of the acid molecules that is ionized is 29%

We have

HClO2 = 0.1 mole = 0.012

HClO + H2O ⇄ ClO₂⁻ + H₂O

Ka = 0.1α x 0.1α / 0.1(1-α)

= 0.1α²/1-α = 0.12

We have to form a quadratic equation from the above

0.1α² = 0.012 - 0.12α

We write the equation to be

0.1α² + 0.012 - 0.12α = 0

To solve the equation above, the quadratic formula calculator was used.

$\frac{-b+-\sqrt{b^2-4ac} }{2}$

alpha = 0.29

This is 29 percent. Therefore the percentage of ionization is 29%.

<h3>What is ionization?</h3>

This is the process through which neutral ions are chraged to become electrically charged.

Read more on ionizations here: brainly.com/question/1445179

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