Question

A certain indicator, HA, has a Ka value of 2.0 x107. The protonated form of the indicator is blue and the ionized form is yellow What is the pKa of the indicator? Number What is the color of this indicator in a solution with pH 5? O blue O yellow O green

Answer 1

$pK_a=-log_{10}K_a$

$pK_a=-log_{10}2.0 \times 10^{-7}$

$pK_a=6.69$

When pH is 5,

Now $10^{-5} > 2.0 \times 10^{-7}$ or ....(1)

and $K_a=[H^+] \times \frac{[A^-]}{[HA]}$ ......(2)

From (1) and (2), we can say that $\frac{[A^-]}{[HA]}<1$

or

Since protonated form of indicator is blue, the colour of the indicator will be blue.