Question

Using the Ka value of 1.4×10−5, place Al(H2O)63+ in the correct location in Figure

Relative acid strength "H20+ HClO, HE H,Coz HCIONH, 2- HPO,2- acids Weaker H0 acids 10-14 Ka 1.0 10-2 104 105 108 10-10 10-12 Relative conjugate base strength heker H20clo, - Ko 10-14 10-12 F 10-20 HCO, 100 CONH 100 101 PO, 10² BH Stronge 10

Answer 1

It will be between HF and H2CO3

As the given value of Ka comes between these two acids Ka ,

For conjugate base,

As. Ka*Kb = 10-14

So, Kb = 7.1×01-10

on second scale ,it will be between, F- and ClO2-