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 

What are DOUBLE SALTS ?

DOUBLE SALTS:

(1)The addition compounds formed by the combination of two simple salts are termed as double salts.

(2) Double salts are stable in solid state only.

(3) When dissolved in water, it furnishes all the ions present in the simple salt form which it has been constituted.

(4)The solution of double salt shows the properties of the samples salts from which it has been constituted

For examples

Mohar’s salt-FeSO₄ (NH₄)₂SO₄ .6H₂O (Ferrous ammonium sulphate)

Alum’s- K₂SO₄Al₂ (SO4)₃. 24H₂O (Potassium ammonium sulphate)

Karnalite- KCl.MgCl_2.6 (H₂O)

Dolomite- CaCO₃.MgCO₃ or CaMg (CO₃)₂