Question

If we consider the limiting molar conductance of the ions; ${H^ + }$to be 344 and $CH_3COO$to be 40, and molar conductance of $0.008M\;CH_3COOH$to be $48$; then determine the value of acid dissociation constant $Ka$of $CH_3COOH$?
(a) $1.4 \times {10^{ - 5}}$
(b) $1.2 \times {10^{ - 5}}$
(c) $1.4 \times {10^{ - 4}}$
(d) $1 \times {10^{ - 5}}$

Answer 1

Hint : When an acid is put into a solution, it starts to dissociate, the amount of dissociation is known as its strength and $Ka$ represents this acid dissociation constant. To find that we can try these steps; first we can determine the degree of dissociation using given data, and since molarity and molar conductance of $CH_3COOH$ is given, we can use the formula to find $Ka$.

Complete Step By Step Answer:
We can first write down the given data and what we need to find.
Given that;
Limiting molar conductance for the constituent ions are as follows;
$\lambda {m^\infty }({H^ + }) = 344$, $\lambda {m^\infty }(CH_3CO{O^ - }) = 40$
Also give the molarity $C = 0.008M\;CH_3COOH$
And said that molar conductance $\lambda {m^C} = 48$
To find: $Ka$ $= ?$
So now we can find the limiting molar conductance of the acid;
$\Rightarrow \lambda {m^\infty }(CH_3COOH) = \lambda {m^\infty }(CH_3CO{O^ - }) + \lambda {m^\infty }({H^ + })$
$\Rightarrow \lambda {m^\infty }(CH_3COOH) = 40 + 344$
$\Rightarrow \lambda {m^\infty }(CH_3COOH) = 384$
The step we need to do is to find the degree of dissociation of acid $(\alpha )$;
$\Rightarrow \alpha = \dfrac{{\lambda {m^C}}}{{\lambda {m^\infty }}}$
$\Rightarrow \alpha = \dfrac{{48}}{{384}}$
$\Rightarrow \alpha = 0.125$
Now we move on to find the dissociation constant;
$Ka = \dfrac{{(C\alpha ) \times (C\alpha )}}{{C(1 - \alpha )}} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }}$
Substituting the values of $C$and $\alpha$;
$\Rightarrow Ka = \dfrac{{0.008 \times {{(0.125)}^2}}}{{1 - 0.125}}$
$\Rightarrow Ka = \dfrac{{0.000125}}{{0.875}}$
$\Rightarrow Ka = 1.4 \times {10^{ - 4}}$
Hence dissociation constant is $1.4 \times {10^{ - 4}}$.
So the option with the correct answer is;
Option (C) $1.4 \times {10^{ - 4}}$ .

Note :
Acids are extremely reactive substances and one should be very careful while handling it. It is said that we should never add water to any acid but in order to dissolve acid in water we must add acid to the water. This is because when water and acids are dissolved, they produce an immense amount of heat which can cause severe burns. So if we put water into acid, since acid is greater in ratio, it will boil quickly and will splash out. But another way around is safe dissolution due to a greater amount of water in the ratio.