Question

At what pH will a ${10}^{-4}M$ solution of an indicator with $K_b=1\times {10}^{-11}$ change colour?

• A. $7.0$
• B. $3.0$
• C. $5.5$
• D. $11.0$

Answer 1

The correct option is B $3.0$

The equilibrium reaction for the basic indicator $\left(InOH\right)$ is given below:

$InOH\rightleftharpoons I{n}^{+}+O{H}^{-}$

The expression for the dissociation constant is $K_b=\frac{\left[I{n}^{+}\right]\left[O{H}^{-}\right]}{\left[InOH\right]}$

When $\left[I{n}^{+}\right]=\left[InOH\right]$, the indicator will change its color.

In such a case,

$K_b=\left[O{H}^{-}\right]=1\times {10}^{-11}$

Hence, $\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1\times {10}^{-14}}{1\times {10}^{-11}}=1\times {10}^{-3}$

Hence, $pH=-log(1\times {10}^{-3})=3$

Hence, the correct option is $B$