Question

Consider an indicator $HIn$. Find the change in $pH$ i.e. $\Delta pH$ of an acidic indicator when the ratio $\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ changes from $0.2$ to $2$.
Given $K_{In}=1\times {10}^{-5}$

• A. $0.2$
• B. $1$
• C. $0.1$
• D. $2$

Answer 1

The correct option is B $1$

$K_{in}=1\times {10}^{-5}p{K}_{in}=-log\left(K_{in}\right)p{K}_{in}=-log(1\times {10}^{-5})p{K}_{in}=5$
$pH=p{K}_{in}+\log \left(\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\right)$
Case 1 :
$p{H}_1=5+log\left(0.2\right)=5-0.7=4.3p{H}_1=4.3$

Case 2 :
$p{H}_2=5+log\left(2\right)=5+0.3=5.3p{H}_2=5.3$
Change in $pH:$
$\Delta pH=p{H}_2-p{H}_1\Delta pH=5.3-4.3=1$