HIn <-> 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