An acid-base indicator has a `K_(a)` of `3.0 xx 10^(-5)`. The acid form of the indicator is red and the basic form is blue. (a) By how much must the `pH` change in order to change the indicator from `75%` red to `75%` blue?

Explanation

`HIn Leftrightarrow H^+ +In^(-)`
`K_(ln)=K_(a)=([H^+] [In^(+)])/([HIn])=([H^+] ["base"])/(["acid"])`
`[H^+]=K_a (["acid"])/(["base"])`
For 75% red, `[H^+]=((3 xx 10^(-5)) (75))/(25)=9 xx 10^(-5)`
pH =4.05
For 75% blue, `[H^+]=((3 xx 10^(-5)) (25))/(75)=1 xx 10^(-5)`
pH =5
The change in pH =5-4.05=0.95.