Notifications
Clear all
Equilibrium
1
Posts
2
Users
0
Likes
275
Views
0
25/10/2021 12:29 pm
Topic starter
If pKb for fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is
(a) 3.52 × 10-3
(b) 6.75 × 10-4
(c) 5.38 × 10-2
(d) 1.74 × 10-5
1 Answer
0
25/10/2021 12:33 pm
Correct answer: (b) 6.75 × 10-4
Explanation:
Kw = Ka x Kb
Kb = 10-10.83 = 1.48 × 10-11
∴ Ka = \(\frac{K_w}{K_b}\)
= \(\frac{10^{-14}}{1.48 \times 10^{-11}}\)
= 6.75 × 10-4