Question

An acid base indicator has a $K_a$ of ${10}^{-6}$. The acid form of the indicator is red and the basic form is blue. By how much must the $pH$ change in order to change the indicator form $80$% red to $80$% blue?

• A. $\Delta pH=1.2$
• B. $\Delta pH=1.7$
• C. $\Delta pH=3.76$
• D. None of these

Answer 1

Answer: (A)

$\Delta pH=1.2$

For an acid-base indicator, $pH=p{K}_{In}+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

So when indicator is 20% red

$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=\frac{20}{80}$

$p{H}_1=6+log\frac{2}{8}$

So when indicator is 20% blue

$\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=\frac{80}{20}=4$

$p{H}_2=6+log\frac{8}{2}$

So $\Delta pH=2log4=4log2=1.2$