Question

A monoprotic acid in a 0.1 m 0.1m solution ionises to 0.001 % 0.001%. Its ionisation constant is

Answer 1

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Answer 2

Answer: $1.0\times 10^{-11}$

Explanation:

The dissociation reaction for a monoprotic acid is given by:

$HA\rightarrow H^++A^-$

    c              0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

$K_a$ = Ionization constant

$K_a=\frac{[H^+][A^-]}{[HA]}$

${\text {Concentration of}} HA=0.1 M$

$\alpha$ = ionization constant =$0.001\%=\frac{0.001}{100}=0.00001M$

$K_a=\frac{[H^+][A^-]}{[HA]}$

$K_a=\frac{[c\alpha][c\alpha]}{[c-c\alpha]}$

Putting in the values we get:

$K_a=\frac{[0.1\times 0.00001][0.1\times 0.00001][}{[0.1-0.1\times 0.00001]}$

$K_a=1.0\times 10^{-11}$

Thus ionization constant is $1.0\times 10^{-11}$