Question

The pH of monoacidic weak base is 11.4. Calculate its percent dissociation in 0.04 M Solution : Define : Electro chemical series​

Answer 1

Answer:

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 %

-a serial arrangement of metallic elements or ions according to their electrode potentials determined under specified conditions; the order shows the tendency of one metal to reduce the ions of any other metal below it in the series.