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HW : Ch 17 Part 2 (Butters, Titrations, Solubility &Complex lons) -3.4hrs)(52 credits Common-lon Effect on Solubi
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Answer #1

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

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