Question

the dissociation constant of nh4oh is 1.8 into 10 raise to minus 5 calculate its degree of dissociation in 0.01 M solution by log table​

Answer 1

Answer:

As we know, conc. of [OH^ - ] = Calpha = √(Kb × C) So, by putting given values, [OH^ - ] = √(1.8 × 10^-5 × 0.1) = 1.34 × 10^-3