For this problem, we’re being asked to calculate the solubility (in mol/L) of PbCl2(s) in 0.100 M KCl(aq). Since the compounds are ionic compounds, they form ions when dissociating in water. The dissociation of PbCl2 and KCl in water are as follows:
The chloride ion, Cl-, has a charge of –1. Pb is a transition metal.
Since ionic compounds criss-cross charges which becomes subscripts:
• Subscript 2 of Cl must have come from Pb → so the charge of Pb is +2 → Pb2+
• Subscript 1 of Pb must have come from the -1 charge of Cl → Cl-
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl–(aq)
The chloride ion, Cl-, has a charge of –1. Potassium is in Group 1A so it’s charge is +1:
KCl(aq) → K+(aq) + Cl–(aq)
Notice that there is a common ion present, Cl–. The common ion effect states that the solubility of a salt is lower in the presence of a common ion.
We can construct an ICE table for the dissociation of PbCl2. Remember that solids are ignored in the ICE table.
Calculate the molar solubility of PbCl 2 (Ksp = 1.6 x 10-5 at 25 degrees) in a 0.100 M solution of KCl.
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