Question

the ph of monoacidic weak base is 12.7 calculate its percent dissociation in 2×10-2 m solution​

Answer 1

Explanation:

monoacidic weak base is 12.7 calculate its percent dissociation in 2×10-2 m solution

Answer 2

Answer:

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Solution : pOH of the solution is given as :

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 %