Question

the degree of ionisation of 0.5N NH3 at 25 degree centigrade in a solution of ph is 12

Answer 1

Answer:

answer to your question is 2%

Answer 2

Answer:

Solution :

NH4OHCmol L−1C−Cα⇔NH+40Cα+OH−0Cα

pH = 12 means [H+]=10−12or[OH−]=10−2

∴[OH+]=Cα=10−2orα=10−2C=10−20.5=2×10−2or2%.