The pH of at which an acid indicator with Ka is 10^-15 changes colour when indicator concentration 1×10^-5 M is?

SOLUTION:

An indicator with Ka = 10^-5 is solution with pH = 6 Calculate % of indicator in ionised form?

SOLUTION:

For acidic indicator, pH range is 3 to 4.6 calculate the ratio of[ In- ] and H+ for the appearance of solution in a single colour.

SOLUTION:

Given pH range 3.0 to 4.6 so pK_In= (3.0+ 4.6)/2 =3.6

For an indicator pKa is 6 Calculate pH of Solution having this indicator such that 40% indicator molecules remain in ionised form.

SOLUTION:  We know that  

The pH range of a basic indicator is 4 to 6.5 Calculate the dissociation constant of indicator?.

SOLUTION:  pK_In must be midpoint of pH range for acidic indicators and pOH range for basic indicators

The pH range = 4 to 6.5 so pOH range is 10 to 7.5

Hence Pk_In =   (10+7.5) /2 = 8.75