Question

An indicator has a standard ionization constant = 9*10^-9 . The acid colour is yellow and the alkalone colour is red . The yellow is visible when the ratio of yellow to red is 3:1 and red colour predominates when red to yellow ratio is 2:1 . What is the pH range of the indicator .

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{KIn}=9\times {10}^{-9} \\ \mathrm{pKIn}=-\log \mathrm{KIn}=9-\log 9=9-0.954=8.046 \\ \mathrm{For}\mathrm{weak}\mathrm{acid}\mathrm{indicator}, \\ \mathrm{pH}=\mathrm{pKIn}+\log \frac{\left[{\mathrm{In}}^{-}\right]\mathrm{Red}}{\left[\mathrm{HIn}\right]\mathrm{Yellow}} \\ {\mathrm{pH}}_1=\mathrm{pKIn}+\log \frac{1}{3} \\ =8.046-\log 3=8.046-0.477 \\ \mathrm{or},{\mathrm{pH}}_1=7.569 \\ {\mathrm{pH}}_2=\mathrm{pKIn}+\log \frac{2}{1} \\ =8.046+0.301 \\ =8.347 \\ \mathrm{pH}\mathrm{range}\mathrm{of}\mathrm{indicator}\mathrm{is}6.569\mathrm{to}8.347 \\ ∆\mathrm{pH}=8.347-7.569=0.778 \end{gathered}$$