Question

CALCULATE THE PH OF A 0.01M ETHANOIC ACID SOLUTION

Answer 1

Answer:

The PH of $0.01M$ concentration of ethanoic acid is $3.37$·

Explanation:

Ethanoic acid is a weak electrolyte and its value of the dissociation constant $K_{a}$ is $1.8$ × $10^{-5}$·

We want to find out the PH at $0.01M$ concentration of ethanoic acid solution·

The dissociation of ethanoic acid follows $;$

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

Initially the concentration of

   $CH_{3} COOH\ \ =\ \ C$

    $CH_{3}COO^{-}\ \ =\ \ 0\\ H^{+}\ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ 0$

At equilibrium, the concentration of

   $CH_{3} COOH\ \ =\ \ C(1-\alpha )\\CH_{3}COO^{-}\ \ =\ \ C\alpha \\ H^{+}\ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ C\alpha$

where,

   α   $=$  degree of dissociation

The dissociation constant for ethanoic acid,

   $K_{a}\ \ =\ \ \ \frac{\left[\begin{array}{ccc}CH_{3}COO^{-} \end{array}\right]\left[\begin{array}{ccc}H^{+} \end{array}\right] }{\left[\begin{array}{ccc}CH_{3}COOH \end{array}\right]}$

           $=\ \ \frac{C\alpha \ *\ C\alpha }{C(1-\alpha )}$

           $=\ \ \ \frac{C^{2} \alpha ^{2} }{C(1-\alpha )}$

$1-\alpha$ is very small, so we can neglect it·

So,

  $K_{a}\ \ =\ \ \ \frac{C^{2} \alpha ^{2} }{C}$

  $K_{a}$     $=\ \ C\alpha ^{2}$

ie,

       $\alpha \ \ =\ \ \sqrt{\frac{K_{a} }{C} }$

⇒    $\alpha \ \ =\ \ \sqrt{\frac{1.8\ *\ \ 10^{-5} }{0.01} }$

⇒          $=\ \ \sqrt{1.8\ \ *\ \ 10^{-3} }$

⇒          $=\ \ 4.24$ × $10^{-2}$

So we get the value of α·

To find out the PH we want to know the concentration of $H^{+}$·

The concentration of $H^{+}$  can be calculated by

⇒   $[H^{+\ } ]\ \ =\ \ C\alpha$

⇒               $=\ \ 0.01$ × $4.24$ × $10^{-2}$

⇒               $=\ \ 4.24$ × $10^{-4}$M

We know that,

  $PH\ \ =\ \ -log[H^{+} ]$

ie,

    $PH\ \ =\ \ -log[4.24\ *\ 10^{-4}]$

⇒  PH       $=\ \ 3.37$