Question

For the acid indicator thymol blue, pH is 2.0 when half the indicator is in unionised form. Find the % of indicator in unionised form in the solution with $[{{H}^{+}}]\text{ = 4 x 1}{{\text{0}}^{-3}}\text{ M}$.
(A) 25.4%
(B) 27.2%
(C) 28.6%
(D) 29.4%

Answer 1

Hint: Write the dissociation reaction for the acid indicator, thymol blue. Since the indicator is only half ionized, concentration of products and reactants will be numerically equal. We can then substitute these values in the formula mentioned below:
$\text{pH = p}{{\text{K}}_{In}}\text{ + log}\frac{[(\text{conjugate ion) }\!\!]\!\!\text{ }}{[(\text{acid indicator) }\!\!]\!\!\text{ }}$
Where,
pH is potenz of hydrogen of the solution
$\text{p}{{\text{K}}_{In}}$ is dissociation constant for acid indicator

Complete step-by-step answer:
Thymol blue is a brownish-green or reddish-brown powder in crystalline form. The formula name for thymol blue is thymol sulfonephthalein. It is commonly used as a pH indicator.
Thymol blue is insoluble in water. However, it is readily soluble when alcohol is used as the solvent or very dilute alkali solutions as well.
Thymol blue changes its color from red to yellow between the range of pH 1.2 - 2.8. Then the color changes from yellow to blue between 8.0 - 9.6 pH values. It is an important component of universal indicators.
We are now going to calculate the % of indicator in unionised form.
The equilibrium reaction for the dissociation of the indicator (HIn) is :
$\text{HIn }\to \text{ I}{{\text{n}}^{-}}\text{ + }{{\text{H}}^{\text{+}}}$
The expression for pH as mentioned in the hint is :
$\text{pH = p}{{\text{K}}_{In}}\text{ + log}\frac{[(\text{conjugate ion) }\!\!]\!\!\text{ }}{[(\text{acid indicator) }\!\!]\!\!\text{ }}$
$\text{pH = p}{{\text{K}}_{In}}\text{ + log}\frac{[I{{n}^{-}}\text{ }\!\!]\!\!\text{ }}{[HIn\text{ }\!\!]\!\!\text{ }}$
2.0 = $\text{p}{{\text{K}}_{In}}$ + $\text{log }\frac{1}{1}$
$\text{p}{{\text{K}}_{In}}$ = 2.0
For $[{{H}^{+}}]\text{ = 4 x 1}{{\text{0}}^{-3}}\text{ M}$, we will calculate the pH
pH = $-\log [{{\text{H}}^{+}}]$
pH = $-\log (\text{4 x 1}{{\text{0}}^{-3}})$ = 2.4
Substituting the values, we get
$2.4\text{ = 2}\text{.0 + log}\frac{[\text{I}{{\text{n}}^{-}}]}{[\text{HIn }\!\!]\!\!\text{ }}\text{ = }\frac{100}{39.8}$
The % of indicator in unionized form in the solution = $\frac{39.8}{100+39.8}\text{ x 100}$ = 28.6%.
Therefore, the correct answer is option (C).

Note: The formula name for thymol blue is thymol sulfonephthalein. The structure is given below for reference: