When the 0.10 m solution of hclo2 is allowed to come to equilibrium, what percent of the acid molecules are ionized?
1 answer:
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.
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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By rewriting the equation from step 1 substituting the value of k into it, the value of x is 22.5
A direct variation is a type of variation that has two types of variables in which a variable can be expressed in terms of the other with its constant.
For example;
If x varies directly as y
<h3>Step 1:</h3>Then
- x
y
- x = ky ; where k is constant
The original question says:
- if x varies directly as y, and x = 9 when y = 6, find x when y = 15
∴
x = ky
9 = k × (6)
k = 9/6
k = 1.5
<h3>Step 3:</h3>Now, we can rewrite the equation from step 1 substituting the value of k into it:
∴
x = ky
- when y = 15 and k = 1.5
The value of (x) will be:
x = 1.5 (15)
x = 22.5
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