if ionisation constant of an acid (HA) equilibrium is 1.0 into 10^-8 then calculate the value of PKa and pkb be for this conjugate base
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pka = 8 and pkb 6 ..
pka = -logka = -log10^-8
= 8 .
and pkw = pka + pkb
14 = 8 + pkb
so pkb = 6.
pka = -logka = -log10^-8
= 8 .
and pkw = pka + pkb
14 = 8 + pkb
so pkb = 6.
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It is said ionisation constant of an acid that is ka
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