At 298 K, the Ka for the dissociation of ethanoic acid in water is 1.8 x 10-5 mol dm-3 and the autoprotolysis constant (Kw) of water is 1.0 x 10-14 mol2 dm-6. Calculate the basicity constant (Kb) of the ethanoate ion in water at 298 K.
I calculated it as;
Ka x Kb = Kw
Kb = Kw/Ka
= 1.0x10-14 / 1.8x10-5 = 5.55x10-10
Is this right?
Thanks
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