# molar solubility

(noun)

## Definition of molar solubility

the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated

Source: Wikipedia - CC BY-SA 3.0

## Examples of molar solubility in the following topics:

• ### Molar Solubility and Relative Solubility

• Molar solubility, which is directly related to the solubility product, is the number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
• The units are molarity, or mole liter-1 (mol/L).The relation between the molar solubility and the solubility product means that one can be used to find the other.Example 1: The Ksp for AgI is 8.5 \cdot 10-17 at 25 °C.
• How would we find the molar solubility?
• Now, solve for "s":s2 = 8.5 x 10-17s = (8.5 x 10-17)1/2s = 9.0 x 10-9 mol/LThe molar solubility of AgI is 9.0 x 10-9 mol/L.Example 2: The solubility products for cadmium carbonate, CdCO3, and silver carbonate, Ag2CO3, are almost exactly the same.
• The solubility is equal to s mol/L.
• Molar solubility is the number of moles of a solute that can be dissolved per liter of solution before the solution becomes saturated.
• ### Complex Ion Equilibria and Solubility

• The absorbance, A, of the complex is proportional to its concentration, M, and can be measured directly on the spectrophotometer: $A=kM$The Beer-Lambert law relates the amount of light being absorbed to the concentration of the substance absorbing the light and the path length through which the light passes:$A = εbc$In this equation, the measured absorbance (A) is related to the molar absorptivity constant ( ε), the path length (b), and the molar concentration (c) of the absorbing species.
• The concentration is directly proportional to absorbance and can be estimated from calibration curve obtained experimentally.Complex Ion Formation and SolubilityFormation of a chemical complex has an effect on solubility.
• $AgCl(s)+2NH3(aq)\rightarrow [Ag(NH_{ 3 })_{ 2 }]^{ + }(aq)+Cl^{ - }(aq)$The equilibrium constant for this reaction is:${ K }_{ c }=\frac { [Ag(NH_{ 3 })_{ 2 }^{ + }][{ Cl }^{ - }] }{ [NH_{ 3 }]^{ 2 } }$The molar solubility of AgCl (x) in 2M NH3 can be calculated knowing that Kc=Kf*Ksp, where Kf = 1.7*107 is the constant of formation of [Ag(NH3)2]+  and Ksp = 1.77*10-10 is solubility product constant for AgCl:${ K }_{ c }=\frac { { x }^{ 2 } }{ [2-2x]^{ 2 } } =Kf*Ksp=1.7\times 10^{ 7 }*1.77×10^{ −10 }$From this, x = 0.006M
• Formation of a complex ion between metal cation and a ligand can increase salt solubility.