Part A
Solubility equilibrium of Pb(SCN)2
Pb(SCN)2(s) <-------> Pb2+(aq) + 2SCN-(aq)
Ksp = [Pb2+][SCN-]2= 2.00 ×10-5
If solubility of Pb(SCN)2 is represented by S
at saturated solution
[Pb2+] = 1S
[SCN-] = 2S
so,
(1S)(2S)2 = 2.00 ×10-5
4S3 = 2.00 × 10-5
S3 = 5.00 × 10-6
S = 0.0171M
Therefore,
Molar solubility of Pb(SCN)2 in pure water = 0.0171M
Part B
Ksp = [ Pb2+] [SCN-]2 = 2.00 × 10-5
substitute given concentration of SCN- in Ksp expression
[Pb2+]( 0.100M)2 = 2.00 ×10-5M3
[Pb2+] = 2.00 × 10-3M
Therefore
molar solubility of Pb(SCN)2 in 0.100M KSCN = 2.00 ×10-3M
HW : Ch 17 Part 2 (Butters, Titrations, Solubility &amp;amp;Complex lons) -3.4hrs)(52 credits Common-lon Effect on...
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