11HIn <--> H+ + In-
Ka =[ H+][in-]/[HIn]
when [In-]/[Hin]=10 ; Ka=[H+] x 10 => [H+]=10-6
pH=6.
Similarly when [In-]/[HIn] = 1/10 pH=4.
therefore min. change =2.
An acid type indicator, HIn differs in colour from its conjugate base (In–). The human eye is sensitive to colour differences only when the ratio [In–] / [HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change
(ka = 1.0 × 10–5)?
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2 Answers
106HIn <-> H+ + In-
Ka = [H+][In-]/[HIn]
taking log on both sides,
pKa = pH - log([In-]/[HIn])
pH = pKa + log([In-]/[HIn])
Let [In-]/[HIn] was 0.1 before pH change and it became 10 after pH change. This is the min change in the ratio for observing a colour change sensitive to human eye
ΔpH = 5 + log(0.1) - (5+log(10))
= 5 -1 -5-1
= -2
So there should be a pH change of atleast 2
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