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Question: The concentration of acetic acid \( \left( {{K_a} = 1.8 \times {{10}^{ - 5}}} \right) \) required to...

The concentration of acetic acid (Ka=1.8×10−5)\left( {{K_a} = 1.8 \times {{10}^{ - 5}}} \right) required to give 3.5×10−4molL−13.5 \times {10^{ - 4}}mol{L^{ - 1}} H+{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}} \\\

Explanation

Solution

The most significant carboxylic acid is acetic acid (CH3COOH)\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.

Complete answer:
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+{H^ + } ion and the anion is CH3COO−C{H_3}CO{O^ - } ion. This statement can be represented as CH3COOH⇌CH3COO−+H+C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + }
The concentration of acetic acid can be found by using the formula:
Ka=[H+][CH3COO−][CH3COOH]{K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {C{H_3}CO{O^ - }} \right]}}{{\left[ {C{H_3}COOH} \right]}}
Since 11 mole of acetic acid produces 11 mole of CH3COO−C{H_3}CO{O^ - } ions and 11 mole of H+{H^ + } ions. We can write this as [CH3COO−]=[H+]\left[ {C{H_3}CO{O^ - }} \right] = \left[ {{H^ + }} \right]
Since 11 mole of CH3COO−C{H_3}CO{O^ - } ions is equivalent to 11 mole of H+{H^ + } ions, we can substitute CH3COO−C{H_3}CO{O^ - } with H+{H^ + } .
After substitution we get,
Ka=[H+]2[CH3COOH]......(1){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, Ka=1.8×10−5{K_a} = 1.8 \times {10^{ - 5}} , and the concentration of hydrogen ions, H+=3.5×10−4molL−1{H^ + } = 3.5 \times {10^{ - 4}}mol{L^{ - 1}}
Substituting the above values in the equation (1)\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 6.8×10−3molL−16.8 \times {10^{ - 3}}mol{L^{ - 1}} . The correct option is B)6.8×10−3molL−1B)6.8 \times {10^{ - 3}}mol{L^{ - 1}} .

Additional Information:
Acetic acid is used to make metal acetates, which are used in some printing processes; vinyl acetate, which is used to make plastics; cellulose acetate, which is used to make textiles and photographic films; and volatile organic esters (such as ethyl and butyl acetates), which are commonly used as solvents for paints, resins, and lacquers.

Note:
If not utilised properly and safely, acetic acid can be a dangerous chemical. Because this liquid is extremely corrosive to the skin and eyes, it must be handled with extreme caution. If consumed or inhaled as vapour, acetic acid can harm the internal organs.