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ehidna [41]
3 years ago
5

What is the percent ionization of a monoprotic weak acid solution that is 0.188 M? The acid-dissociation (or ionization) constan

t, Ka, of this acid is 2.43 × 10 − 12 .
Chemistry
1 answer:
djyliett [7]3 years ago
5 0

Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is 3.59\times 10^{-4}\%

Explanation:

Dissociation of weak acid is represented as:

HA\rightleftharpoons H^+A^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.188 M and \alpha = ?

K_a=2.43\times 10^{-12}

Putting in the values we get:

2.43\times 10^{-12}=\frac{(0.188\times \alpha)^2}{(0.188-0.188\times \alpha)}

(\alpha)=3.59\times 10^{-6}=3.59\times 10^{-4}\%

Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is 3.59\times 10^{-4}\%

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