Question

A certain indicator, HA, has a K, value of 0.00040. The protonated form of the indicator is yellow and the ionized form is red. What is the pK, of the indicator? PK = What is the color of this indicator in a solution with pH = 5? O yellow orange red

Answer 1

The given solution has Ka = 0.00040

So pka= -log (ka)

= -log (0.00040)

= 3.39

So the pka value is 3.39

At pH above the pka the functional group is deprotonated meaning it is in the ionic form

And color of indicator in ionized form is red

So the answer is

01. pKa= 3.39

02. Red

I hope this helps. If you have any query or want more detailed explanation, feel free to ask in the comments section below.