The indicator dinitrophenol is an acid with a \({K_a}\) of \(1.1 \times 1{0^{ - 4}}\). In a \(1.0 \times 1{0^{ - 4}} - M\) solution, it is colourless in acid and yellow in the base. Calculate the \(pH\) range over which it goes from \(10\% \) ionized (colourless) to \(90\% \) ionized (yellow).

The reaction

-

The value of \({K_a}\) of the indicator dinitrophenol is \(1.1 \times 1{0^{ - 4}}\)

- It is colourless in acid and yellow in base

- Let Hin represent the dinitrophenol molecule, which is colourless.

- Let \(l{n^ - }\) represent the deprotonated dinitrophenol molecule, which is yellow.

Let us calculate the \(pH\) range over which the indicator dinitrophenol goes from \(10\% \) ionized (colourless) to \(90\% \) ionized (yellow).

The equilibrium in a solution of the indicator dinitrophenol, which is a weak acid:

The equilibrium expression

\({K_a} = \frac{{\left[ {{C_6}{H_3}\left( {N{O_2}} \right){O^ - }} \right] \times \left[ {{H_3}{O^ + }} \right]}}{{\left[ {{C_6}{H_3}{{\left( {N{O_2}} \right)}_2}OH} \right]}}\) (10\%)

First, let us calculate the \(pH\) when \(10\% \) of the dinitrophenol is ionised.

- The concentration of the deprotonated dinitrophenol (base) is

\(\begin{array}{l}1.0 \times 1{0^{ - 4}}:100\% = x:10\% \\x = 1.0 \times 1{0^{ - 5}}M\end{array}\)

- The concentration of the dinitrophenol as an acid is

\(1.0 \times 1{0^{ - 4}} - 1.0 \times 1{0^{ - 5}} = 9.0 \times 1{0^{ - 5}}M\)

- The concentration of hydronium ion is

\(\begin{array}{l}{K_a} = \frac{{\left[ {I{n^ - } \times \left[ {{H_3}{O^ + }} \right]} \right.}}{{[HIn]}}\\\left[ {{H_3}{O^ + }} \right] = \frac{{{K_a} \times [HIn]}}{{\left[ {I{n^ - }} \right]}}\\ = \frac{{1.1 \times 1{0^{ - 4}} \times 9.0 \times 1{0^{ - 5}}}}{{1.0 \times 1{0^{ - 5}}}}\\ = 9.9 \times 1{0^{ - 4}}M\end{array}\)

And the\(pH\)of a solution is

\(\begin{array}{c}{\bf{pH = - log}}\left( {{{\bf{H}}_{\bf{3}}}{{\bf{O}}^{\bf{ + }}}} \right)\\{\bf{ = - log}}\left( {{\bf{9}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}} \right)\\{\bf{ = 3}}{\bf{.00}}\end{array}\)

\((90\% )\)

Now, let us calculate the\(pH\)when\(90\% \)of dinitrophenol is ionized.

- The concentration of the deprotonated dinitrophenol (base) is

\(\begin{array}{l}1.0 \times 1{0^{ - 4}}:100\% = x:90\% \\x = 9.0 \times 1{0^{ - 5}}M\end{array}\)

- The concentration of the dinitrophenol as an acid is

\(1.0 \times 1{0^{ - 4}} - 9.0 \times 1{0^{ - 5}} = 1.0 \times 1{0^{ - 5}}M\)

- The concentration of hydronium ion is

\(\begin{array}{c}{K_a} = \frac{{\left[ {I{n^ - } \times \left[ {{H_3}{O^ + }} \right]} \right.}}{{[HIn]}}\\\left[ {{H_3}{O^ + }} \right] = \frac{{{K_a} \times [HIn]}}{{\left[ { In{ ^ - }} \right]}}\\ = \frac{{1.1 \times 1{0^{ - 4}} \times 1.0 \times 1{0^{ - 5}}}}{{9.0 \times 1{0^{ - 5}}}}\\ = 1.2 \times 1{0^{ - 5}}M\end{array}\)

\(\)And the\(pH\)of a solution is

\(\begin{array}{c}{\bf{pH = - log}}\left( {{{\bf{H}}_{\bf{3}}}{{\bf{O}}^{\bf{ + }}}} \right)\\{\bf{ = - log}}\left( {{\bf{1}}{\bf{.2 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}} \right)\\{\bf{ = 4}}{\bf{.92}}\end{array}\)

Therefore, the pH range is from\(3.00 to 4.92\)