Question

Bromophenol blue is an acidic indicator with Ka value of $6\times {{10}^{-5}}$. What
percentage of this indicator is in its basic form at a pH of 5? (Given that log 6 =0.78)
a.) 82.5%
b.) 85.7%
c.) 83.6%
d.) 87.2%

Answer 1

Hint: In a dissociation reaction which is also called dissociation constant, the liberated proton combines with a water molecule to give a hydronium ion, the dissociation should be written as acid-base reaction.
$HA+{{H}_{2}}O⇌{{A}^{-}}+{{H}_{3}}{{O}^{+}}$
acid + base ⇌ conjugate base + conjugate acid.

Complete step by step solution:
The acid dissociation constant or the ionization constant is the measure of the strength of an acid in the solution. Its equilibrium constant for a reaction
HA = ${{H}^{+}}$+${{A}^{-}}$
The chemical species HA is an acid which dissociates as ${{A}^{-}}$, the conjugate base of the acid.
The system is said to be in equilibrium, when the concentration of its components does not change with time, as both forward and backward reactions are occurring at the same time.
Let's write the chemical equation:
$H{{B}_{b}}(aq)+{{H}_{2}}O(l)={{B}_{b}}^{-}+{{H}_{3}}{{O}^{+}}(aq)$
Let's take the total acidic solution ($H{{B}_{b}}$) as 100, and the basic (${{B}_{b}}^{-}$) as x.
So, the concentration of $H{{B}_{b}}$ will be 100-x.
Now we know,
pH= -log [${{H}_{3}}{{O}^{+}}$]
5 = -log [${{H}_{3}}{{O}^{+}}$]
[${{H}_{3}}{{O}^{+}}$] = ${{10}^{-5}}$
${{K}_{a}}$ = [${{B}_{b}}^{-}$] $\dfrac{[{{H}_{3}}{{O}^{+}}]}{[H{{B}_{b}}]}$
6 × ${{10}^{-5}}$ = x × $\dfrac{{{10}^{-5}}}{100-x}$
$\dfrac{x}{100-x}$= 6
600 - 6x = x
600 = 7x
x = 600/7
x = 85.7 %

Therefore, from the above solution we can conclude that the correct answer is (b).

Note: The acid dissociation constant for an acid is a direct consequence of the underlying thermodynamics of the dissociation constant. The $p{{K}_{a}}$ value is directly proportional to the standard Gibbs energy change in the reaction.