Question

The ionization constant of acetic acid is 1.74 x 10⁻⁵. 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

Answer:

The degree of dissociation of acetic acid $\alpha = 1.86\times 10^{-2}$

The concentration of acetate ion in the solution, $[H^{+}] =9.33 \times 10^{-4}$

pH value is = 3.03

Explanation:

As per the question,

Given data:

Ionization constant of acetic acid, $K_{a}=1.74 \times 10^{-5}$

$CH_{3}COOH = c = 0.05\ M$

As we know that Ionization of acetic acid gives,

$CH_{3}COOH \Rightarrow CH_{3}COO^{-} + H^{+}$

By using the formula:

$k_{a} = \frac{Concentraton\ of\ products}{Concentration\ of\ rectants}$

$k_{a} = \frac{[CH_{3}COO^{-}] +[H^{+}]}{[CH_{3}COOH]}$

As we know that concentration of $[CH_{3}COO^{-}]=[H^{+}]$

So after putting the values,we can write,

$1.74 \times 10^{-5} = \frac{[H^{+}] +[H^{+}]}{0.05}$

After solving, we get

$\Rightarrow[H^{+}] =9.33 \times 10^{-4}$

Therefore,

The concentration of acetate ion in the solution, $[H^{+}] =9.33 \times 10^{-4}$

Now,

To calculate pH:

pH = -log[H⁺]

     = -log[9.33 × 10⁻⁴]

     = 3.03

Hence, pH value is = 3.03

Now for degree of dissociation:

$\alpha =\sqrt{\frac{k_{a}}{c}}$

$\alpha =\sqrt{\frac{1.74 \times 10^{-5}}{0.05}}$

$\alpha = 1.86\times 10^{-2}$

Therefore,

The degree of dissociation of acetic acid $\alpha = 1.86\times 10^{-2}$