If pKb for the fluoride ion at 25∘C is 10.83 The ionization constant – InterviewSolution

1.	If pKb for the fluoride ion at 25∘C is 10.83 The ionization constant of hydrofluoric acid in water at this temperature will be: (Given:10−10.83=1.48×10−11)
Answer» If $p K_{b}$ for the fluoride ion at $25^{\circ} C$ is 10.83 The ionization constant of hydrofluoric acid in water at this temperature will be: $(G i v e n : 10^{- 10.83} = 1.48 \times 10^{- 11})$

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