Question

The common-ion effect can be applied to a solution of lead thiocyanate when potassium thiocyanate is added. Since there is a common ion, the cyanate ion, the molar solubility of lead thiocyanate goes down in the presence of potassium thiocyanate. Calculate the molar solubility of lead thiocyanate in 0.600 M KSCN.

Answer 1

Answer:

5.56 ×10⁻⁵M

Explanation:

The formula for lead thiocyanate = Pb(SCN)₂ and for  potassium thiocyanate, we have KSCN.

The standard solubility product $K_{sp$ of Pb(SCN)₂ is known to be 2.0 × 10⁻⁵

Since there are 0.600 M of KSCN, definitely if KSCN dissociates we have:

KSCN         ⇄         K⁺       +        SCN⁻

0.600M               0.600 M          0.600 M

On the other hand, if Pb(SCN)₂ dissociates ; we have:

Pb(SCN)₂          ⇄         Pb⁺       +        2SCN⁻

x                                     x                     0.600 M

$K_{sp$  = x(0.600)²

2.0 × 10⁻⁵ = x (0.36)

x = $\frac{2.0*10^{-5}}{0.36}$

x = 5.56 × 10⁻⁵ M

∴ The Molar Solubility of lead thiocyanate in 0.600 M KSCN = 5.56 × 10⁻⁵ M