Question

Ionization constant of${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$is$1.7 \times {10^{ - 5}}$and concentration of${{\rm{H}}^{\rm{ + }}}$ions is$3.4 \times {10^{ - 4}}$. Then find out initial concentration of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$molecules:
A.$3.4 \times {10^{ - 4}}$
B.$3.4 \times {10^{ - 3}}$
C.$6.8 \times {10^{ - 4}}$
D.$6.8 \times {10^{ - 3}}$

Answer 1

Hint: We know that the product of the concentration of hydrogen ion and the anion when divided by the concentration of undissociated acid, then the ratio is constantly termed as the dissociation constant of that acid. This constant is characteristic of every acid, and this constant varies only with temperature.

Step by step answer:
From the question, it is given that
The ionisation constant of that${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$is$1.7 \times {10^{ - 5}}$.
The concentration of $\left[ {{{\rm{H}}^{\rm{ + }}}} \right]$ ions is $3.4 \times {10^{ - 4}}$.
Let us first derive the formula, from the law of chemical equilibrium, the equilibrium constant ${{\rm{K}}_{\rm{C}}}$is given by the expression,
${{\rm{K}}_{\rm{C}}} = \dfrac{{\left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]}}$

Water is present in excess quantity in dilute solutions, so we will take its concentration may be taken constant known as k. we know that hydronium ion indicates the hydrogen ion so it can be replaced by $\left[ {{{\rm{H}}^{\rm{ + }}}} \right]$
So, our equation becomes,

${{\rm{K}}_{\rm{C}}} = \dfrac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]\left[ {\rm{k}} \right]}}$
The product of the two constants is another constant known as ${K_a}$.So the equation becomes,

${{\rm{K}}_{\rm{a}}} = \dfrac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}$
The ionisation reaction of acetic acid is shown below.

${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}} \to {\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - } + {{\rm{H}}^{\rm{ + }}}$
The ionisation constant of acetic acid is shown below.

${\rm{k}} = \dfrac{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }} \right]\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right]}}$
Where, k is the ionisation constant.

At equilibrium x the value of $\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }} \right] = \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 3.4 \times {10^{ - 4}}$.
Substitute the respective values in the above equation.
$\begin{array}{c}
1.7 \times {10^{ - 5}} = \dfrac{{\left( {3.4 \times {{10}^{ - 4}}} \right)\left( {3.4 \times {{10}^{ - 4}}} \right)}}{{\rm{c}}}\\
{\rm{c}} = 6.8 \times {10^{ - 3}}\,{\rm{M}}
\end{array}$
Where, c is the concentration of acetic acid.
Thus, the concentration of acetic acid is \[6.8 \times {10^{ - 3}}\,{\rm{M}}\].

Hence, the correct option for this question is C.

Note: The degree of dissociation of an acid is generally used to measure its capacity which produces the hydrogen ions and so it is also used to measure the strength of the substances.