Notifications
Clear all
Equilibrium
1
Posts
2
Users
0
Likes
322
Views
0
25/10/2021 12:38 pm
Topic starter
If Ksp(PbSO4) = 1.8 × 10-8 and Ka(HSO4-) = 1.0 × 10-2 the equilibrium constant for the reaction.
PbSO4(s) + H+(aq) ⇌ HSO4-(aq) + Pb2+(aq) is
(a) 1.8 × 10-6
(b) 1.8 × 10-10
(c) 2.8 × 10-10
(d) 1.0 × 10-2
1 Answer
0
25/10/2021 12:45 pm
Correct answer: (a) 1.8 × 10-6
Explanation:
PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)(Ksp) .....(i)
HSO4-(aq) ⇌ H+(aq) + SO42-(aq)(Ka) .....(ii)
Subtracting equation (ii) from (i), then
PbSO4(s) + H+(aq) ⇌ Pb2+(aq) + HSO4-(aq)(Keq)
∴ Keq = \(\frac{K_{sp}}{K_a}\)
= \(\frac{1.8 \times 10^{-8}}{1.0 \times 10^{-2}}\) = 1.8 × 10-6