Question

The ionization constant of acetic acid is $1.74\text{ }\times \text{ 1}{{\text{0}}^{-5}}$. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer 1

Hint: The degree of dissociation is defined as the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. To find the ionization we need to know the equation that is mentioned in the question. Then we can find the pH of the solution.

Complete step-by-step answer:
It is given that,
The ionization constant of acetic acid is $1.74\text{ }\times \text{ 1}{{\text{0}}^{-5}}$ and concentration is 0.05M.
The ionization of acetic acid is:
$\text{C}{{\text{H}}_{\text{3}}}\text{COOH }\overset{{}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}}\text{ + }{{\text{H}}^{\text{+}}}$
Therefore,
${{\text{K}}_{\text{a}}}\text{ = }\dfrac{\left[ \text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}} \right]\left[ {{\text{H}}^{\text{+}}} \right]}{\left[ \text{C}{{\text{H}}_{\text{3}}}\text{COOH} \right]}\text{ = }\dfrac{{{\left[ {{\text{H}}^{\text{+}}} \right]}^{2}}}{\left[ \text{C}{{\text{H}}_{\text{3}}}\text{COOH} \right]}$
$\therefore \text{ }\left[ {{\text{H}}^{+}} \right]\text{ = }\sqrt{\left( 1.74\text{ }\times \text{ 1}{{\text{0}}^{-5}} \right)\left( 5\text{ }\times \text{ 1}{{\text{0}}^{-2}} \right)}\text{ = 9}\text{.33 }\times \text{ 1}{{\text{0}}^{-4}}\text{ = }\left[ \text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}} \right]$

${{\text{K}}_{\text{a}}}$ is known as the acid dissociation constant.
So the pH of the solution is
$\begin{align}
  & =\text{ }-\log \left( 9.33\text{ }\times \text{ 1}{{\text{0}}^{-4}} \right) \\
 & =\text{ 4 }-\text{ log(9}\text{.33)} \\
 & \text{= 3}\text{.03} \\
\end{align}$
We know that,
$\text{ }\!\!\alpha\!\!\text{ = }\sqrt{\dfrac{{{\text{K}}_{\text{a}}}}{\text{C}}}$
$\text{ }\!\!\alpha\!\!\text{ = }\sqrt{\dfrac{1.74\text{ }\times \text{ 1}{{\text{0}}^{-5}}}{\text{C}}}\text{ = 1}\text{.86 }\times \text{ 1}{{\text{0}}^{-2}}$

Additional Information:
In case of solution, when we say equilibrium, we mean that there are no changes in concentration of products and reactants over time.

Ionization constant is also known as the measure of the acid strength. The higher will be the value of the ${{\text{K}}_{\text{a}}}$, the greater will be the number of hydrogen ions liberated per mole of acid in solution. And thus stronger will be the acid.

Note: To find the pH of an aqueous solution, we need to know the concentration of the hydronium ion in moles per litre. This is also known as molarity.
By ionization constant we mean a constant which depends upon the equilibrium between the ions and the molecules that are not ionized in a solution or liquid.