English, asked by aagerwal8693, 1 month ago

The pH of monoacidic weak base is11.2 calculate it's present dissociation in0.02m solution

Answers

Answered by anveshadeshmukh68
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 %

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