The pH of monoacidic weak base is11.2 calculate it's present dissociation in0.02m solution
Answers
Answered by
0
Solution= pOH of the solution is given
pOH=14-pH=14-11.2=2.8
pOH= -log10[OH]
log10[OH]= -pOH
=-2.8= -2-0.8-1+1
=-3+0.2=3.2
[OH]= antilog 3.2= 1.585× 10-3 mol/dm3
For monoacidic base,
BOH (aq) B⊕(aq) + OH (aq)
[OH ] = ∝c
∝ = [OH ]c
= 1.585 × 10-3
0.02 = 0.07925
Percent dissociation = ∝ × 100
= 0.07925 × 100
= 7.925 %
Similar questions