Question

Compare the acidic nature of $0.1M$ $HCNO$ with that of $0.2M$ $HCN$ . Given ${K_a}$ for
$HCNO = 1.2 \times {10^{ - 4}}$ and ${K_a}$ for $HCN = 4.0 \times {10^{ - 10}}$ at $298K$.
A . $HCNO$ solution is $2.6 \times {10^{ - 3}}$ times more acidic
B . $HCNO$ solution is $3 \times {10^5}$ times more acidic
C . $HCNO$ solution is $3.9 \times {10^2}$ times more acidic
D . none of these$HCNO \rightleftharpoons {H^ + } + CN{O^ - }$

Answer 1

- Hint: Lets approach this question with the help of the concept of chemical equilibrium where the rate of reversible reactions at which the concentration of the reactants and products do not change with time is known as chemical equilibrium. According to the law of mass action,”Rate of reaction is directly proportional to the raised to the power equal to the respective stoichiometric coefficient ”. Let’s take an example of a reaction that is where P and Q reacts and gives R and S as product and this is an reversible reaction,
                $pP + qQ \rightleftarrows rR + sS$
Rate of forward reaction $ \propto {[P]^p}{[Q]^q}$
Rate of forward reaction \[ = {k_f}{[P]^p}{[Q]^q}\]
Rate of backward reaction \[ \propto {[R]^r}{[S]^s}\]
Rate of backward reaction \[ = {k_b}{[R]^r}{[S]^s}\]
At equilibrium , ${k_f}{[P]^p}{[Q]^q} = {k_b}{[R]^r}{[S]^s}$
                               ${K_a} = \dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{[P]}^p}{{[Q]}^q}}}{{{{[R]}^r}{{[S]}^s}}}$
Where ${K_a} = $ equilibrium constant,${k_f} = $rate constant for forward reaction and $k{_b} = $rate constant for backward reaction.

Complete step-by-step solution -
Now we will first write an equation where the following product is formed at equilibrium and we will try to find out the concentration of $[{H^ + }]$ ion which will show the acidic nature of $HCNO$.
$HCNO \rightleftharpoons {H^ + } + CN{O^ - }$
In the problem concentration of $HCNO$ and $HCN$ is given which is respectively $0.1M$ and $0.2M$. $ \Rightarrow {K_a} = \dfrac{{[{H^ + }][CN{O^ - }]}}{{[HCNO]}}$
$ \Rightarrow 1.2 \times {10^{ - 4}} = \dfrac{{[{H^ + }][CN{O^ - }]}}{{0.1}}$
$(\because [{H^ + }] = [CN{O^ - }])$
$[{H^ + }] ^{2} =0.12 \times {10^{ - 4}}$
So $[{H^ + }] = 0.35 \times {10^{ - 2}}M$
Again we will do the same procedure which we have done in above reaction , $HCN \rightleftharpoons {H^ + } + C{N^ - }$
$ \Rightarrow {K_a} = \dfrac{{{{[{H^ + }]}^2}}}{{[HCN]}}$
$ \Rightarrow 4 \times {10^{ - 10}} = \dfrac{{{{[{H^ + }]}^2}}}{{[HCN]}}$
$(\because [{H^ + }] = [C{N^ - }])$
$ [{H^ + }]^{2} = 0.8 \times {10^{ - 10}}$
So $[{H^ + }] = 0.89 \times {10^{ - 5}}$ M
That means $HCNO$ is $3.9 \times {10^2}$ = $\left( {\dfrac{{0.35 \times {{10}^{ - 2}}}}{{0.89 \times {{10}^{ - 5}}}}} \right)$ times more acidic than $HCN$

Note : We have approached this problem by applying the concept of equilibrium ,in the problem concentration of $HCNO$ and $HCN$ is given .With the help of mass action law we got the concentration $[{H^ + }] = 0.35 \times {10^{ - 2}}M$ when we consider reaction $HCNO \rightleftharpoons {H^ + } + CN{O^ - }$ and again with the same procedure we got concentration $[{H^ + }] = 0.89 \times {10^{ - 5}}$ M and then we compared both concentration of acidic ion.