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# The ionization constant of acetic acid is$\text{1}\text{.8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-5}}$. The concentration at which it will be dissociated to $2{\scriptstyle{}^{0}/{}_{0}}$ is: A) $\text{1M}$ B) $\text{0}\text{.045M}$ C) $\text{0}\text{.018M}$ D) $\text{0}\text{.45M}$

Last updated date: 15th Jun 2024
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Hint: The weak acids do not undergo the complete dissociation. The dissociation constant or ionization constant ${{\text{K}}_{\text{a}}}$is directly proportional to the square of the degree of dissociation $(\alpha )$ and concentration of the solution. Thus substitute the values and get the unknown quantity.

Complete step by step solution:
We know that the acetic acid when dissolved in water does not undergo the complete dissociation. Here we are given that it is partially dissociated into the water. The acetic acid has dissociated to $2{\scriptstyle{}^{0}/{}_{0}}$ of its total concentration.
We are provided with the ionization constant ${{\text{K}}_{\text{a}}}$ of the acid.
${{\text{K}}_{\text{a}}}\text{= 1}\text{.8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-5}}$
The acid ionization constant i.e. ${{\text{K}}_{\text{a}}}$is the equilibrium constant between the ionized and unionized acid. It represents the fraction in which the undissociated acid ionized into the solution. It reflects the strength of the acid.
We know that the acetic acid undergoes the incomplete dissociation. It dissociates into the acetate ion $\text{CH3CO}{{\text{O}}^{-}}$and proton${{\text{H}}^{\text{+}}}$. Since acetic acid is weak acid the equilibrium constant is always towards the reactant side. The acetate ion and proton can combine to again regenerate the acetic acid. Therefore, the dissociation of acid as shown below,
$\begin{matrix} {} & \text{C}{{\text{H}}_{\text{3}}}\text{COOH} & \rightleftarrows & \text{CH3CO}{{\text{O}}^{-}} & \text{+} \\ \text{Before dissociation} & \text{C} & {} & \text{0} & {} \\ \text{After dissociation} & \text{(1- }\!\!\alpha\!\!\text{ )C} & {} & \text{ }\!\!\alpha\!\!\text{ C} & {} \\ \end{matrix}\begin{matrix} \text{ }{{\text{H}}^{\text{+}}} \\ \text{ 0} \\ \text{ }\!\!\alpha\!\!\text{ C} \\ \end{matrix}$
The ionization constant can be written as,
\begin{align} & \text{Ka = }\dfrac{\text{Conc}\text{. of dissociated species}}{\text{Con}\text{.of undissociated specie}}\text{ } \\ & \text{ = }\dfrac{\left( \text{ }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\alpha\!\!\text{ C} \right)}{\left( \text{1- }\!\!\alpha\!\!\text{ } \right)\text{}} \\ & \therefore \text{Ka= }\dfrac{\left( {{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} \right)}{\left( \text{1- }\!\!\alpha\!\!\text{ } \right)}\text{C} \\ \end{align}
The value of α is very small compared to 1. That is $\alpha \ll 1$ thus we neglect the term α from the denominator. We get,
$\text{Ka= }{{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}}\text{C}$
We know that the acetic acid dissociates to the $2{\scriptstyle{}^{0}/{}_{0}}$. Therefore, the degree of dissociation $(\alpha )$ will be:
$\alpha =\dfrac{2}{100}=\text{ 0}\text{.02}$
We are interested to find the value of the concentration of acetic acid. Thus rearrange the equation of the ionization constant for concentration. We have,
\begin{align} & \text{Ka= }{{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}}\text{C} \\ & \Rightarrow \text{C = }\dfrac{\text{Ka}}{{{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}}} \\ \end{align}
Let us substitute the values in the rearranged equation. We get,
\begin{align} & \text{C = }\dfrac{\text{Ka}}{{{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}}} \\ & \text{C}=\dfrac{1.8\times {{10}^{-5}}}{{{\left( 0.02 \right)}^{2}}} \\ & C=\dfrac{1.8\times {{10}^{-5}}}{4\times {{10}^{-4}}} \\ & \therefore \text{C = 0}\text{.045 M} \\ \end{align}
Therefore the concentration of acetic acid is$\text{0}\text{.0467 M}$.

Hence, (B) is the correct option.

Note: The degree of dissociation for a weak acid is very small compared to the unity thus it is always negligible. Similarly, it is applicable for weak bases. Remember that such types of questions can be extended to calculate the pH value of acid.