Question

1.0 g of a weak monobasic acid (HA), when dissolved in 150 ml water, lowers the freezing point by ${{0.186}^{0}}C$. Also 1.0 g of the same acid required 125 ml of a 0.10N Nate solution for complete neutralisation. Determine dissociation constant (${{K}_{a}}$) of the weak acid. ${{K}_{f}}$ of water is 1.86 K kg $mo{{l}^{-1}}$ .
A.$4.16\times {{10}^{-3}}$
B.$2.38\times {{10}^{-6}}$
C.$6.43\times {{10}^{-3}}$
D.$3.78\times {{10}^{-6}}$

Answer 1

Hint: Try to find out the expected molar mass of the acid by using the depression in freezing point and the weights. Also, find the normal molar mass using the equivalent concept. Now relate both the molar masses with the degree of dissociation.

Complete answer:
In order to solve the question, let us first know what dissociation constant actually is. Dissociation constant can be said as a special type of equilibrium constant that shows us the ability of a particular salt, its ability to dissociate into its constituent ions. Now, write down the values we have in our question. We have
${{W}_{2}}=1g$(acid weight)
${{W}_{1}}=150g$, weight of water, assuming density=1
$\Delta {{T}_{f}}=0.168$, depression in freezing point
So, On applying the formula of depression in freezing point, we get the value of ${{M}_{\exp }}$ of acid i.e expected molar mass of acid.
$\begin{align}
  & \Delta {{T}_{f}}={{K}_{f}}\times \dfrac{{{W}_{2}}}{{{M}_{2}}}\times \dfrac{1000}{{{W}_{1}}} \\
 & \\
\end{align}$
So,
$0.168=\dfrac{1.86\times 1\times 1000}{{{M}_{\exp }}\times 150}$
On solving, ${{M}_{\exp }}=73.80$……….( i )
Now, using the concept of equivalence, we have
${{M}_{eq}}acid={{M}_{eq}}alkali$
On solving, we have
$\dfrac{1}{{{E}_{w}}}\times 1000=125\times \dfrac{1}{10}$
On solving, we obtain
${{E}_{w}}acid=80$
As 1g of monobasic acid is used so,
Normal molecular mass ${{M}_{n}}=80\times 1=80$
Now, let us see how the acid is getting dissociated. Let us assume $\alpha $ as degree of dissociation:
$\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HA\Leftrightarrow {{H}^{+}}+{{A}^{-}} \\
 & at\,t=0,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0 \\
 & at\,t={{t}^{'\,\,}},\,\,\,1-\alpha \,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\alpha \\
\end{align}$
So, the total concentration is $1-\alpha +\alpha +\alpha =1+\alpha $. Now,
$\dfrac{{{M}_{n}}}{{{M}_{\exp }}}=1+\alpha $, so
$\Rightarrow \dfrac{80}{73.80}=1+\alpha $
On solving, $\alpha =0.2$, and the concentration is $\dfrac{1}{8}\times \dfrac{100}{15}=0.083$. Finally, we have
$\begin{align}
  & {{K}_{a}}=\dfrac{c{{\alpha }^{2}}}{1-\alpha } \\
 & =\dfrac{0.083\times {{(0.2)}^{2}}}{0.8} \\
 & =4.16\times {{10}^{-3}} \\
\end{align}$

So we get the value of ${{K}_{a}}$ to be $4.16\times {{10}^{-3}}$, which gives us option A as the correct answer.

Note:
The degree of dissociation and percentage dissociation are somewhat similar. Percentage dissociation can be obtained by multiplying by the degree of dissociation by 100. Also, degree of dissociation is inversely proportional to the square root of the concentration.