Test: Solubility Product - NEET MCQ

Ba₃(PO₄)₂ ⇌ 3Ba⁺²  (2PO₄)^-3

K_sp =  [Ba⁺²>]³ [(PO₄)^-3]²
[3s]³[2s]²
K_sp = 108s⁵
Or s = (K_sp/108)^⅕

A general equation of equilibrium constant for a reaction of dissociation of PbI₂ as PbI₂⇌Pb²⁺+2I^-can be written as,

But we know that K_sp is the equilibrium constant at the saturated level. At saturation, no more PbI₂ will dissolve. Thus the concentration of PbI₂ will be constant and can be taken as 1.
Therefore the above equation can be written as,

For precipitation, Qrkn > Ksp, then tto establish equilibrium, ions will combine to give molecule as ppt.
So applying the above concept,
 1.8×10^-10 < [Ag⁺][Cl^-]
 1.8×10^-10 < 10^-5 × [Cl^-]
Or [Cl-] > 1.8×10^-5
By seeing the option, only option d satisfies this condtion.

Molar solubility(s) for PbBr₂ =  2.17×10^-3 M
PbBr₂   ⇌   Pb⁺² + 2Br^-
s                      s          2s
K_sp = [Pb⁺²][Br^-]²
= [s][s]²
= 4s³  =  4(2.17×10^-3)³
= 4.1×10^-8

The expression for the solubility product is Ksp =

The solubility product is the product of ionic concentrations raised to appropriate stoichiometric coefficients.

PbCl₂ ⇌ Pb²⁺ + 2Cl^– or, K_sp = [Pb²⁺][Cl^–]²     

The correct answer is Option B.
BaCl₂ ⇔ Ba²⁺ + 2Cl⁻
Ksp = (s)(2s)² = 4s³
Given, 4s³ = 10⁻⁶
⇒s³ = 0.25 ⨯ 10⁻⁶
⇒s = (0.25)³ ⨯ 10⁻²
⇒s = 0.629 ⨯ 10⁻²
⇒s = 6.3 ⨯ 10⁻³

In saturated solution of NaCl,equillibrium is established between ions and unionised NaCl. When HCl gas is passed through saturated solution of NaCl,it is completely dissociated to form H+ and Cl- ions. Cl- ion is common to both electrolyte ,hence Cl- ion concentration in solution increases.

Definition of Solubility Product, K.A substance's solubility product, Ksp, is the mathematical product of its dissolved ion concentrations raised to the power of their stoichiometric coefficients. Solubility products are relevant when a sparingly soluble ionic compound releases ions into solution.

The correct answer is Option C.
The ionization equilibrium of silver (I) sulfide is 
Ag₂S⇌2Ag⁺ +S2⁻
The solubility product expression for silver (I) sulfide is KSP[Ag⁺]2 [S2⁻].
But [Ag⁺]² = x and [S²⁻] = y.
Hence the expression for the solubility product becomes K_SP[Ag⁺]² [S²⁻] = x²y.

The salt with more K_sp is more soluble in water. Hence BaCrO₄ is more soluble.

PRECIPITATION: Precipitation occurs when tiny droplets of water, ice, or frozen water vapor join together into masses too big to be held above the earth. They then fall to the ground as precipitation.

A salt is soluble if it dissolves in water to give a solution with a concentration of at least 0.1 moles per liter at room temperature.