Question

Bromophenol blue is an indicator with a $K_a$ value of $6\times {10}^{-5}$. What % of this indicator is in its basic form at a $pH$ of $5$?

• A. $14.29$%
• B. $54.82$%
• C. $78.48$%
• D. $85.71$%

Answer 1

The correct option is D $85.71$%

As we know, $pH=p{K}_{In}+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

$5=-log[6\times {10}^{-5}]+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

$log6=log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=6$

So % of indicator in its basic form $=\frac{6}{7}\times 100=85.71$