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.

5.56 ×10⁻⁵M

Explanation:

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

The standard solubility product 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

 = x(0.600)²

2.0 × 10⁻⁵ = x (0.36)

x =

x = 5.56 × 10⁻⁵ M

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

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in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm