Question

For phenolphthalein $p{K}_{in}$ = 9.7. The indicator can be used for the pH range of 8.2 to 10. Calculate the $\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ for the upper and lower limits of the range.

• A. 1.99
• B. 4.5
• C. 6.3
• D. 8.8

Answer 1

The correct option is A

1.99

$HIn\rightleftharpoons H^{+}+I{n}^{-}$

pH = $p{K}_{in}$ + log$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

For Lower limit:

8.2 = 9.7 + log$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ $\Rightarrow$ log$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ = -1.5

$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ = 0.03162

For Upper limit:

10 = 9.7 + log$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ $\Rightarrow$ log$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ = 0.3

$\therefore \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ = 1.995