The number of moles of a solute that may be dissolved in one liter of a solution is known as molar solubility. It’s measured in milligrams per liter (mol/L) or milligrams per liter (M (molarity). And what is the molar solubility of pbbr2 in a 0.500 m pb(no3)2 solution?
The molar concentrations of lead (II) ions and bromide ions in a saturated solution of lead (II) bromide are extremely tiny, and seem trivial when compared to the molar concentration of lead (II) ions in a 0.500 M solution of lead (II) nitrate, based on the value of Ksp constant.
By substituting 0.500 M for the lead (II) ions concentration in the equation for the Ksp constant of lead (II) bromide, we can determine the molar concentration of bromide ions in this solution:
Ksp = [Pb^2+][(2)(Br^-)]^2
6.60 × 10^-6 = [0.500][(2)(Br^-)]^2
6.60 × 10^-6 = [0.500](4)[Br^-]^2
13.2 × 10^-6 = (4)[Br^-]^2
3.30 × 10^-6 = [Br^-]^2
1.8165902 × 10^-6 = [Br^-]
Because the concentration of lead (II) ions from dissolved lead (II) bromide in this solution will be half that of the bromide ions:
[Pb^2+] = (0.5)(1.8165902 × 10^-6) = 0.908296 × 10^-6 mol/L or
[PbBr2] = 9.08 × 10^-7 mol/L
The molar solubility of PbBr2 in 0.500 M Pb(NO3)2 solution is approximately 9.08 × 10^-7 mol/L.
Lead(II) bromide, 367.01 g/mol / Molar mass
To determine solubility in mole/L, multiply the number of moles by the volume of the solution in liters. The volume of the solution in our case is 55 mL (0.055 L). NaNO3 has a solubility of 0.258 moles/0.055 L, or 4.69 mole/L.
The main distinction between molar solubility and product solubility constant is that the former explains the dissolving of a material per litre of a solution, whilst the latter describes the dissolution of a solid substance in an aqueous solution.
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