Question

An acid type indicator, Hln differs in colour from its conjugate base ($I{n}^{-}$). The human eye is sensitive to colour differences only when the ratio $\frac{\left[I{n}^{-}\right]}{\left[Hln\right]}$ is greater than 10 or smaller than 0.1. What would be the minimum change in pH of the solution to observe a complete colour change ($K_a$ = $1.0\times {10}^{-5}$) ?

• A. 2
• B. 3.25
• C. 6
• D. 4

Answer 1

The correct option is A

2

Here, to register an observable change the ratio $(\frac{\left[I{n}^{-}\right]}{\left[Hln\right]}$ should be greater than 10 or lesser than 10-1.Irrespective of how we effect the change, the magnitude should cross either barrier.

For case I

$pH$ = $p{K}_a$ + $log\frac{\left[salt\right]}{\left[Acid\right]}$ = $log(1.0\times {10}^{-5})+log\left(10\right)$ = $5+1$ = $6$

For case II

$pH$ = $p{K}_a$ + $log\frac{\left[salt\right]}{\left[Acid\right]}$ = $log(1.0\times {10}^{-5})$ + $log\left(0.1\right)$ = $5$ - $1$ = $4$

If we inflict a change in $pH$ which is lesser than 6 - 4 = 2, then there is a chance that the human eye might not register it. Thus we need to at the very least, alter the $pH$ by 2.