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Question: What is the \[\left[ {{H}^{+}} \right]\]in a \(0.40M\) solution of \(HOCl\), \({{K}_{a}}=3.5\times {...

What is the [H+]\left[ {{H}^{+}} \right]in a 0.40M0.40M solution of HOClHOCl, Ka=3.5×10−8{{K}_{a}}=3.5\times {{10}^{-8}} ?
A.1.4×10−8M1.4\times {{10}^{-8}}M
B.1.2×10−4M1.2\times {{10}^{-4}}M
C.1.9×10−9M1.9\times {{10}^{-9}}M
D.3.7×10−4M3.7\times {{10}^{-4}}M

Explanation

Solution

Equilibrium is defined as a state of a system where the concentration of the reactant and the concentration of product do not change with respect to time. At equilibrium, there will be no change in the properties. The rate of forward reaction is equal to the rate of backward reaction.

Complete step-by-step answer: Let us see the dissociation reaction of hypochlorous acid is-
HOCl(aq)→H+(aq)+OCl−(aq)HOCl\left( aq \right)\to {{H}^{+}}\left( aq \right)+OC{{l}^{-}}\left( aq \right)
The dissociation constant of acid can be written as
Ka=[H+][OCl−][HOCl]{{K}_{a}}=\dfrac{\left[ {{H}^{+}} \right]\left[ OC{{l}^{-}} \right]}{\left[ HOCl \right]}
In the starting of the reaction, 0.40M0.40Mof HOClHOClwas there and no moles of H+{{H}^{+}} and OCl−OC{{l}^{-}} ions. But as the system reaches equilibrium, xx moles of HOClHOClgets dissociated. Let us see the chemical reaction:
HOCl(aq)→H+(aq)+OCl−(aq)HOCl\left( aq \right)\to {{H}^{+}}\left( aq \right)+OC{{l}^{-}}\left( aq \right)
t=0t=0 0.400.40 00 00
At equilibrium 0.40−x0.40-x xx xx
In this question, it is given that the value of dissociation constant of an acid (Ka)\left( {{K}_{a}} \right) is 3.5×10−83.5\times {{10}^{-8}}
On substituting the value in the formula of dissociation constant of acid we get,
3.5×10−8=[x][x][0.40−x]3.5\times {{10}^{-8}}=\dfrac{\left[ x \right]\left[ x \right]}{\left[ 0.40-x \right]}
We can neglect the value of x in the denominator as the dissociation constant of an acid is very small.
3.5×10−8=[x][x][0.40]3.5\times {{10}^{-8}}=\dfrac{\left[ x \right]\left[ x \right]}{\left[ 0.40 \right]}
x2=[3.5×10−8][0.40]{{x}^{2}}=\left[ 3.5\times {{10}^{-8}} \right]\left[ 0.40 \right]
On further solving we get,
x=1.9×10−9Mx=1.9\times {{10}^{-9}}M
Here, [H+]=x\left[ {{H}^{+}} \right]=x
Hence, [H+]=1.9×10−9M\left[ {{H}^{+}} \right]=1.9\times {{10}^{-9}}M

Therefore, the correct option is C.

Note: There are factors that affect equilibrium:
Change in concentration: when there is a change in the concentration of the reactant and the concentration of the product then there will be a change in the composition of equilibrium.
Change in temperature: in an exothermic reaction, when the temperature is increased then equilibrium constant decreases but in endothermic reaction, when the temperature is increased then equilibrium constant increases.
Effect of catalyst: there is no effect of catalyst in equilibrium. Catalyst equally speeds up the forward and backward reaction.