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Question: A weak acid,\[{\text{HA}}\] has \[{K_a}\]of \[{\text{1}}{\text{.00}} \times {\text{1}}{{\text{0}}^{ ...

A weak acid,HA{\text{HA}} has Ka{K_a}of 1.00×10−5{\text{1}}{\text{.00}} \times {\text{1}}{{\text{0}}^{ - 5}}. If 0.100 mole of this acid is dissolved in 1 L{\text{1}}\,{\text{L}}of water. The percentage of acid dissociated at equilibrium is closest to:
A.{\text{99}}{\text{.0% }}
B.{\text{1}}{\text{.00% }}
C.{\text{99}}{\text{.9% }}
D.{\text{0}}{\text{.100% }}

Explanation

Solution

Based on the dissociation of the acids in an aqueous solution there are two types of them that are strong and weak.
A strong acid is one that dissociates or ionizes completely in the solutions while weak acid is one that is not completely dissociated into solutions.
Therefore, in the case of the weak acids, their extent of dissociation is given in terms of the degree of dissociation.

Complete step-by-step answer: Here, in the question, it is given that 0.1 moles of the acid are dissolved in one liter of the water it indicates the concentration of the acid is 0.1 M{\text{0}}{\text{.1}}\,{\text{M}}.
Here, the acid given is weak therefore, its dissociation is given using the degree of dissociation that is alphaalpha .
The weak acid dissociates as follows:
     HA    ⇄       H +         +          A−\,\,\,\,\,{\text{HA}}\,\,\,\, \rightleftarrows \,\,\,\,\,\,\,{{\text{H}}^{\text{ + }}}\,\,\,\,\,\,\,{\text{ + }}\,\,\,\,\,\,\,\,\,{{\text{A}}^ - }
initial            0.1             0           0{\text{initial}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}}{\text{.1}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}}\,\,\,\,\,\,\,\,\,\,\,{\text{0}}
at time ’t’   0.1(1−α)        0.1α         0.1α{\text{at time '}}t{\text{'}}\,\,\,0.1\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,0.1\alpha \,\,\,\,\,\,\,\,\,0.1\alpha
The dissociation constant of the acid is represented as follows:

Ka = [H + ][A−][HA]{K_a}{\text{ = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}
⇒Ka=0.1α ×0.1α 0.1(1−α) \Rightarrow {K_a} = \dfrac{{0.1\alpha \, \times 0.1\alpha \,}}{{0.1\left( {1 - \alpha } \right)\,}}
⇒Ka=(0.1α)2 0.1(1−α) \Rightarrow {K_a} = \dfrac{{{{\left( {0.1\alpha } \right)}^2}\,}}{{0.1\left( {1 - \alpha } \right)\,}}
Here, it is a weak acid therefore the value ofα\alpha is very less than the one. therefore, (1−α)\left( {1 - \alpha } \right) is taken as 1.
Ka=(0.1α)2 0.1(1−α) {K_a} = \dfrac{{{{\left( {0.1\alpha } \right)}^2}\,}}{{0.1\left( {1 - \alpha } \right)\,}}
⇒Ka=(0.1α)2 0.1 \Rightarrow {K_a} = \dfrac{{{{\left( {0.1\alpha } \right)}^2}\,}}{{0.1\,}}
⇒Ka=0.1α2\Rightarrow {K_a} = 0.1{\alpha ^2}
Now, substitute 1.00×10−5{\text{1}}{\text{.00}} \times {\text{1}}{{\text{0}}^{ - 5}} for Ka{K_a}.
⇒1.00×10−5=0.1α2\Rightarrow 1.00 \times {10^{ - 5}} = 0.1{\alpha ^2}
⇒α2=1.00×10−4\Rightarrow {\alpha ^2} = 1.00 \times {10^{ - 4}}
⇒α=0.01\Rightarrow \alpha = 0.01
In terms of the percentage, it is given as follows:
α=0.01×100%=1%\alpha = 0.01 \times 100\% = 1\%
Thus, the percentage of acid dissociated is 1.

Note: The molarity is used to express the concentration of the solutions. It is expressed in units of molar(M) that is mole per liter. It is calculated by taking the ratio of the moles of the solute to the volume of the solution in liters.

The dissociation constant is the measure of the strength of the acid. The higher the value of the dissociation constant stronger is the acid and vice versa. It is represented as Ka{K_a} for acids.
In the case of the weak acids, the value of the dissociation constant is less.