Consider the solution containing 0.181 m lead ions and 0.174
mmercury (ii) ions. the ksp for lead sulfide
is3.4*10-28 and that for mercury (ii) sulfide
is4.0*10-53. calculate the concentration in (m)
ofsulfide ions required to reduce the concentration of mercury
(ii)ions to 1.0*10-6 m?

[ S2- ] = 4.0 E-47 M

Explanation:

∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]

∴ [Pb2+] = 0.181 M

∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]

∴ [Hg2+] = 0.174 M

∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:

∴ [ Hg2+ ] = 1.0 E-6 M

⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M

acid rain does not usually kill trees directly. instead, it is more likely to weaken the trees by damaging their leaves, limiting the nutrients available to them, or poisoning them with toxic substances slowly released from the soil. the main atmospheric pollutants that affect trees are nitrates and sulphates.

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