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Question: Compare the acidic nature of \(0.1M\) \(HCNO\) with that of \(0.2M\) \(HCN\) . Given \({K_a}\) for ...

Compare the acidic nature of 0.1M0.1M HCNOHCNO with that of 0.2M0.2M HCNHCN . Given Ka{K_a} for
HCNO=1.2×10−4HCNO = 1.2 \times {10^{ - 4}} and Ka{K_a} for HCN=4.0×10−10HCN = 4.0 \times {10^{ - 10}} at 298K298K.
A . HCNOHCNO solution is 2.6×10−32.6 \times {10^{ - 3}} times more acidic
B . HCNOHCNO solution is 3×1053 \times {10^5} times more acidic
C . HCNOHCNO solution is 3.9×1023.9 \times {10^2} times more acidic
D . none of theseHCNO⇌H++CNO−HCNO \rightleftharpoons {H^ + } + CN{O^ - }

Explanation

Solution

- 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⇄rR+sSpP + qQ \rightleftarrows rR + sS
Rate of forward reaction ∝[P]p[Q]q \propto {[P]^p}{[Q]^q}
Rate of forward reaction =kf[P]p[Q]q = {k_f}{[P]^p}{[Q]^q}
Rate of backward reaction ∝[R]r[S]s \propto {[R]^r}{[S]^s}
Rate of backward reaction =kb[R]r[S]s = {k_b}{[R]^r}{[S]^s}
At equilibrium , kf[P]p[Q]q=kb[R]r[S]s{k_f}{[P]^p}{[Q]^q} = {k_b}{[R]^r}{[S]^s}
Ka=kfkb=[P]p[Q]q[R]r[S]s{K_a} = \dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{[P]}^p}{{[Q]}^q}}}{{{{[R]}^r}{{[S]}^s}}}
Where Ka={K_a} = equilibrium constant,kf={k_f} = rate constant for forward reaction and kb=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+][{H^ + }] ion which will show the acidic nature of HCNOHCNO.
HCNO⇌H++CNO−HCNO \rightleftharpoons {H^ + } + CN{O^ - }
In the problem concentration of HCNOHCNO and HCNHCN is given which is respectively 0.1M0.1M and 0.2M0.2M. ⇒Ka=[H+][CNO−][HCNO] \Rightarrow {K_a} = \dfrac{{[{H^ + }][CN{O^ - }]}}{{[HCNO]}}
⇒1.2×10−4=[H+][CNO−]0.1\Rightarrow 1.2 \times {10^{ - 4}} = \dfrac{{[{H^ + }][CN{O^ - }]}}{{0.1}}
(∵[H+]=[CNO−])(\because [{H^ + }] = [CN{O^ - }])
[H+]2=0.12×10−4[{H^ + }] ^{2} =0.12 \times {10^{ - 4}}
So [H+]=0.35×10−2M[{H^ + }] = 0.35 \times {10^{ - 2}}M
Again we will do the same procedure which we have done in above reaction , HCN⇌H++CN−HCN \rightleftharpoons {H^ + } + C{N^ - }
⇒Ka=[H+]2[HCN]\Rightarrow {K_a} = \dfrac{{{{[{H^ + }]}^2}}}{{[HCN]}}
⇒4×10−10=[H+]2[HCN]\Rightarrow 4 \times {10^{ - 10}} = \dfrac{{{{[{H^ + }]}^2}}}{{[HCN]}}
(∵[H+]=[CN−])(\because [{H^ + }] = [C{N^ - }])
[H+]2=0.8×10−10[{H^ + }]^{2} = 0.8 \times {10^{ - 10}}
So [H+]=0.89×10−5[{H^ + }] = 0.89 \times {10^{ - 5}} M
That means HCNOHCNO is 3.9×1023.9 \times {10^2} = (0.35×10−20.89×10−5)\left( {\dfrac{{0.35 \times {{10}^{ - 2}}}}{{0.89 \times {{10}^{ - 5}}}}} \right) times more acidic than HCNHCN

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