Question

If pKb for fluoride ion at 25 degree celsius is 10.83,the ionisation constant of hydrofluoric acid in water at this temperature is???

please tell me how to do long calculations in this.

Answer 1

we know, $pK_b=-log_{10}K_b$

so, $pK_b=10.83=-log_{10}K_b$

$K_b=10^{-10.83}$

we know, the reaction is ...
HF + H2O <=> F– + H3O+

so, $K_a.K_b=K_w$

$K_a=\frac{K_w}{K_b}$

$K_a=\frac{10^{-14}}{10^{-10.83}}$

[ we know, Kw = 10^-14 ]

$K_a=10^{-14+10.83}$

$K_a=10^{-3.17}$

$K_a=6.76\times10^{-4}$

hence, ionisation constant of hydrofluoric acid is 6.76 × 10^-4

Answer 2

Explanation:

Solution

Hope its help you mate