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0.2 moler solution of formic acid is 3.2% ionized. Its ionization constant is

Answers (1)

As we discussed in concept

Ionization constant of weak acids -

Consider 

HX(aq)+H_{2}O(l)\rightleftharpoons H_{3}O^{+}(aq)+\bar{X}(aq)
 

K_{a}=\text{ dissociation or ionization constant }

- wherein

K_{a}=\frac{C^{2}\:\alpha ^{2}}{C(1-\alpha )}

K_{a}=\frac{C\alpha ^{2}}{1-\alpha }
 

C=\text{ initial concentration of undissociated acid }

\alpha =\text{ extent upto which HX is ionized into ions.}

 

 \\C=0.2M\\\alpha =3.2\% \: or \: 0.032

\underset{c(1-\alpha)}{HCOOH} \rightleftharpoons \underset{c\alpha }{H^+} + \underset{c\alpha }{COOH^-}

 

Ionization constant K_{a}=\:\:\:\:\:\:\:\frac{[COOH^{-})][H^{+}]}{[HCOOH]}

                              K_{a}\:=\:\frac{c\alpha \times c\alpha }{c(1-\alpha )}

                           =   \frac{(0.2\times 0.032)^{2}}{0.2\times 0.968}

                           = 2.1\times 10^{-4}

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lovekush

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