Question

8.
If the value of K, of 1 M HCN is 10-5 then its degree
of dissociation in 0.1 M HCI will be (a <<< 1)
(1) 10-5
(2) 104
(4) 10-2
(3) 10-3​

Answer 1

The value of degree  of dissociation of 1 M $HCN$ in 0.1 M HCI is $10^{-4}$

Explanation:

The dissociation of strong acid is shown by:

$HCl\rightleftharpoons H^+Cl^-$

0.1 moles of HCl will give 0.1 moles of $H^+$

The dissociation of weak acid is shown by:

$HCN\rightleftharpoons H^+CN^-$  initial:  cM              0             0

at eqm: $c-c\alpha$        $c\alpha+0.1$          $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)(c\alpha+0.1)}{c-c\alpha}$

Give c= 1 M and $\alpha$ = ?

$K_a=10^{-5}$

Putting in the values we get:

$10^{-5}=\frac{(1\times \alpha)(1\alpha+0.1)}{(1-1\times \alpha)}$

As $\alpha$  <<<<1 , neglecting  $\alpha$  in comparison to 1

$10^{-5}=\frac{(1\times \alpha)(1\alpha+0.1)}{(1)}$

$(\alpha)=10^{-4}$

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Answer 2

Answer:

10^-4

Explanation:

Ca^2 is very small so we ignore it as a is very small compared to 1. so now we have 0.1a...a is the degree of dissociation which I have taken...

hope it helps!