Question

Calculate the molar solubility of CuX (Ksp=1.27×10−36) in each of the following. pure water

Answer 1

Answer: CuX = Cu2+ + X2- Ksp = [Cu2+] * [X2-] for each mole of CuX that dissolves we get x mol of each of the anions and cations Ksp = x^2 = 1.27 x 10 ^-36 x= 1.13 x 10 ^-18 moles of CuX per liter of pure water if the solution has [Cu2+]= 0.27 M Ksp becomes x ( x + 0.27) as we can see above x is extremely small so can be ignored inside the brackets 0.27 x = 1.27 x 10^-36 x = 1.27 x 10^-36 / 0.27 = 4.70 x 10 ^-36 moles per liter In 0.19M X2- we have Ksp = 0.19x = 1.27 x 10^-36 x = 1.27 x 10^-36 / 0.19 = 6.68 x 10 ^-36 moles per liter

Answer 2

The molar solubility of CuX : 1.127.10⁻¹⁸ mol/L

Further explanation  

Solubility(s) is the maximum amount of a substance that can dissolve in some solvents.  

Ksp is an ion product in equilibrium  

Solubility (s) and solubility constants (Ksp) of the AxBa solution can be stated as follows.  

AₓBₐ (s) ← ⎯⎯⎯⎯ → aAᵃ⁺ (aq) + b Bᵇ⁻ (aq)  

   s                         as                bs  

Ksp = [Aᵃ⁺]ᵃ [Bᵇ⁻]ᵇ

Ksp = (as)ᵃ (bs)ᵇ

Solubility units in the form of mol / liter or gram / liter  

Ksp is the product of ions in an equilibrium saturated state

The molar solubility of CuX in pure water

Reaction

CuX ⇔ Cu²⁺ + X²⁻

s             s          s

Ksp = s²

1.27×10⁻³⁶ = s²

$\rm s=\sqrt{1.27.10^{-36}}\\\\s=1.127.10^{-18}\:\dfrac{mol}{liter}$

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