Question

The initial rate of hydrolysis of methyl acetate $\left( 1M \right)$by a weak acid $\left( HA,1M \right)$ is ${{\left( \dfrac{1}{100} \right)}^{th}}$ of that of a strong acid $\left( HX,1M \right)$ at ${{25}^{\circ }}C$. The ${{K}_{a}}$ of $HA$ is:
A. $1\times {{10}^{-4}}$
B. $1\times {{10}^{-5}}$
C. $1\times {{10}^{-6}}$
D. $1\times {{10}^{-3}}$

Answer 1

Hint: In this question we are provided with the relative comparison between the rates of hydrolysis by a weak and strong acid. We will use this comparison to find out the value of hydronium ions of weak acid. Once hydronium ions are calculated, then we will calculate the degree of dissociation $\left( \alpha \right)$ and once that is calculated then we can easily calculate the value of ${{K}_{a}}$ by using relation ${{K}_{a}}=C{{\alpha }^{2}}$.

Complete step-by-step answer:
Let us firstly understand that what does hydrolysis means in very simple terms.
Hydrolysis for acids in very simple terms means the dissociation of acids into their corresponding ions.
We also know that strong acid has the tendency to dissociate completely in the water. So, we can say that whatever will be the concentration of strong acid, its ions will even have the same concentration. In this question we are given with the strong acid $HX$. Let us see its dissociation.
$\begin{align}
  & HX\left( aq \right)\to {{H}^{+}}\left( aq \right)+{{X}^{-}}\left( aq \right) \\
 & \left( 1M \right)\text{ }\left( 1M \right)\text{ }\left( 1M \right) \\
\end{align}$
This means that the concentration of $\left[ {{H}^{+}} \right]$ions = $1M$
Now, we also know that weak acid dissociates partially in water and hence the concentration of hydronium ion of weak acid will be comparatively lesser as compared with the strong acid.
Therefore, the dissociation of the weak acid can be represented as
$\begin{align}
  & HA\left( aq \right)\to {{H}^{+}}(aq)+{{A}^{-}}(aq) \\
 & \text{(1-}\alpha \text{) }\alpha \text{ }\alpha \\
\end{align}$
So, $\left[ {{H}^{+}} \right]=\alpha $ (for weak acid)
Now, when we talk of the rate of hydrolysis for acids it actually means the dissociation of concentration of reactants (acid) in a second or formation of products (hydronium ion) in one second.
Also, in question a statement is given to us that
Rate of hydrolysis of weak acid = $\dfrac{1}{100}$(rate of the hydrolysis of strong acid)
Or we can say that,
Rate of formation of products (hydronium ion) of weak acid = $\dfrac{1}{100}$ rate of formation of products (hydronium ion) of strong acid.
 $\left[ {{H}^{+}} \right]$ of weak acid = $\dfrac{1}{100}$$\left[ {{H}^{+}} \right]$ of strong acid. (equation 1)
Now we already know that $\left[ {{H}^{+}} \right]$=$\alpha $ for weak acids and $\left[ {{H}^{+}} \right]$=$1M$ for strong acid.
Therefore,
$\alpha $= $\dfrac{1}{100}\times 1$
$\therefore \alpha ={{10}^{-2}}$
Now we know that the dissociation constant of the acid and degree of dissociation are related to each other as follows:
${{K}_{a}}=C{{\alpha }^{2}}$
$\therefore {{K}_{a}}=1\times {{\left( {{10}^{-2}} \right)}^{2}}=1\times {{10}^{-4}}$
Therefore, the value of ${{K}_{a}}=1\times {{10}^{-4}}$ for the weak acid HA.

Hence, the correct option is Option A.

Note: It should be noted that dissociation constant is basically a quantitative measure of acidic strength of an acid in solution. Acid dissociation constants are mostly used for the weak acids as they are the one which are partially dissociated and for the strong acids, acid dissociation constant is generally very large as they dissociate completely into their corresponding ions.