Soln 1 ) GIVEN
conc of (CH3)2NH2Br = 0.432 M
Kb of (CH3)2NH = 5.42 x 10^-4
=> Ka = 10^-14 / Kb
= 1.845 x 10^-11
now , we know that
(CH3)2NH2Br < ---------------------------------------> (CH3)2NH2^+ + Br-
=> (CH3)2NH2^+ + H2O <--------------------------------------> (CH3)2NH + H3O+
Ka = [(CH3)2NH][H3O^+] / [(CH3)2NH2+]
=> 1.84 x 10^-11 = x^2 / 0.432 ( since Ka is very small , 0.432 - x = 0.432)
=> x = 2.82 x 10^-6
=>[H3O^+] = 2.82 X 10^-6 M
we know that ,
pH = - log[H3O^+]
=> pH = 5.55 ANS
Soln 2) GIVEN
conc of KClO = 0.421 M
Ka of HClO = 2.95 x 10^-8
=> Kb = 10^-14 / Ka
= 3.39 x 10^-7
we know that
KClO <-------------------------------> K ^+ + ClO^-
ClO^- + H2O <-------------------------------> HClO + OH^-
Kb = [HClO][OH-] / [ClO-]
=> 3.39 x 10^-7 = x^2 / 0.421 ( since Kb is very small , 0.421 - x = 0.421)
=> x = 3.78 x 10^-4
=>[OH-] = 3.78 x 10^-4
we know that ,
pOH = -log[OH-]
= 3.42
and ,
pH + pOH = 14
=> pH = 10.58 ANS
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