If pK_b for fluoride ion (F^-) at 25,^{circ}C | vedclass

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(C) Given $pK_b = 10.83$ for $F^-$. The base dissociation constant $K_b = 10^{-pK_b} = 10^{-10.83} = 1.479 	imes 10^{-11} \approx 1.48 	imes 10^{-11

Vedclass · Accepted Answer

$6.75 	imes 10^{-4}$

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(C) Given $pK_b = 10.83$ for $F^-$. The base dissociation constant $K_b = 10^{-pK_b} = 10^{-10.83} = 1.479 	imes 10^{-11} \approx 1.48 	imes 10^{-11}$. At $25\,^{\circ}C$,the ionic product of water $K_w = 10^{-14}$. The relationship between $K_a$ and $K_b$ for a conjugate acid-base pair is $K_a 	imes K_b = K_w$. Therefore,$K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{1.48 	imes 10^{-11}} = 6.756 	imes 10^{-4} \approx 6.75 	imes 10^{-4}$.

If $pK_b$ for fluoride ion $(F^-)$ at $25\,^{\circ}C$ is $10.83$,the ionisation constant $(K_a)$ of hydrofluoric acid $(HF)$ in water at this temperature is:

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