Question

An acid-base indictor has $K_b$ of $1.0\times {10}^{-5}$. The acid form of the indicator is red and the basic form is blue. Then:

• A. pH is 8.4 when indicator is 80% red
• B. pH is 9.6 when indicator is 80% blue
• C. pH is 9.6 when indicator is 80% red
• D. ph is 8.4 when indicator is 80% blue

Answer 1

Answer: (A), (B)

The correct options are
A pH is 8.4 when indicator is 80% red
B pH is 9.6 when indicator is 80% blue

For acid-base conjugate pair, $k_a{k}_b=1.0\times {10}^{-14}$
$k_a=1.0\times {10}^{-14}/K_b=1.0\times {10}^{-14}/1.0\times {10}^{-5}=1.0\times {10}^{-9}$
For ionization of acid-base indicator HIn :
$HIn\rightleftharpoons H^{+}+I{n}^{-}{K}_a=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]}pH=p{K}_a+log\frac{\left[I{n}^{-}\right]}{HIn}$
(a) For 8% red form (HIn), pH $=9+log\frac{20}{80}=8.4$
(b) For 80% blue form$\left(I{n}^{-}\right)$, pH $=9+log\frac{20}{80}=9.6$