Given that Pb(SCN)₂, has a Ksp value of 2.00×10⁻⁵

Pb(SCN)₂ <-------------> Pb2+ + 2SCN-

s 2s s = molar solubility of lead thiocyanate

Ksp = [Pb²⁺] [SCN-]²

Given that [KSCN] = [SCN-] = 1 M

Hence,

Ksp = [s] [2s + 1 M ]²

2.00 × 10⁻⁵ = 4s³ + 4s² + s

Since s is very small,

s³ and s² can be neglected.

Then,

2.00×10⁻⁵ = s

s = 2.00×10⁻⁵ M

Therefore,

molar solubility of lead thiocyanate in 1.00 M KSCN = 2.00×10⁻⁵ M