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The concentration of acetic acid $\left( {{K_a} = 1.8 \times {{10}^{ - 5}}} \right)$ required to give $3.5 \times {10^{ - 4}}mol{L^{ - 1}}$ ${H^ + }$ , is $A)6.8mol{L^{ - 1}} \\ B)6.8 \times {10^{ - 3}}mol{L^{ - 1}} \\ C)1.94 \times {10^{ - 1}}mol{L^{ - 1}} \\ D)194mol{L^{ - 1}} \\$

Last updated date: 13th Jul 2024
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Hint: The most significant carboxylic acid is acetic acid $\left( {C{H_3}COOH} \right)$ , also known as ethanoic acid. Vinegar is a dilute solution of acetic acid made from the fermentation and oxidation of natural carbohydrates; acetate is a salt, ester, or acylal of acetic acid.

A substance's concentration is the amount of solute contained in a given volume of solution. Molarity, or the number of moles of solute in one litre of solution, is commonly used to represent concentrations.
We know that acetic acid dissociates to form two ions in which the cation is ${H^ + }$ ion and the anion is $C{H_3}CO{O^ - }$ ion. This statement can be represented as $C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + }$
The concentration of acetic acid can be found by using the formula:
${K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {C{H_3}CO{O^ - }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$
Since $1$ mole of acetic acid produces $1$ mole of $C{H_3}CO{O^ - }$ ions and $1$ mole of ${H^ + }$ ions. We can write this as $\left[ {C{H_3}CO{O^ - }} \right] = \left[ {{H^ + }} \right]$
Since $1$ mole of $C{H_3}CO{O^ - }$ ions is equivalent to $1$ mole of ${H^ + }$ ions, we can substitute $C{H_3}CO{O^ - }$ with ${H^ + }$ .
After substitution we get,
${K_a} = \dfrac{{{{\left[ {{H^ + }} \right]}^2}}}{{\left[ {C{H_3}COOH} \right]}}......\left( 1 \right)$
From the question, we know that the acid dissociation constant of acetic acid, ${K_a} = 1.8 \times {10^{ - 5}}$ , and the concentration of hydrogen ions, ${H^ + } = 3.5 \times {10^{ - 4}}mol{L^{ - 1}}$
Substituting the above values in the equation $\left( 1 \right)$ , we get
$1.8 \times {10^{ - 5}} = \dfrac{{{{\left[ {3.5 \times {{10}^{ - 4}}} \right]}^2}}}{{\left[ {C{H_3}COOH} \right]}} \\ \left[ {C{H_3}COOH} \right] = \dfrac{{12.25 \times {{10}^{ - 8}}}}{{1.8 \times {{10}^{ - 5}}}} \\ \left[ {C{H_3}COOH} \right] = 6.8 \times {10^{ - 3}}mol{L^{ - 1}} \\$
Therefore, the concentration of acetic acid is equal to $6.8 \times {10^{ - 3}}mol{L^{ - 1}}$ . The correct option is $B)6.8 \times {10^{ - 3}}mol{L^{ - 1}}$ .