Question

Calculate the $pH$ at which a basic indicator $\left(InOH\right)$ with $K_{In}=1\times {10}^{-10}$ changes its colour.
Given : The concentration of indicator is $1\times {10}^{-2}M$.

• A. 8
• B. 4
• C. 2
• D. 10

Answer 1

The correct option is B 4

The colour of an indicator always changes at midpoint , i.e. when it is $50\%$ ionized.

$InOH\left(aq\right)\rightleftharpoons I{n}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
$K_{In}=\frac{\left[O{H}^{-}\right]\left[I{n}^{+}\right]}{\left[InOH\right]}$
When colour changes,
$\left[I{n}^{+}\right]=\left[InOH\right]$
Putting in the above equation we get,
$K_{In}=\left[O{H}^{-}\right]-\log \left(K_{In}\right)=-\log \left[O{H}^{-}\right]$
Since ,
$pOH=p{K}_{In}\left(basic\right)=-\log {10}^{-10}=10pH=14-pOH$ at ${25}^{o}C$
$pH=14-10=4$
note: pH range of the colour change of the indicator does not depend on the concentration of the indicator.