Question

Steps receive zero UE. LULA QUE 2.1. What is the lead concentration of a saturated solution of lead (11) sulfate containing 0,010 M Na SOA? Ksp sou6.3 10.

Answer 1

What is the lead concentration of a saturated solution of lead (II) sulfate containing o.oLoM Na₂ 504? Ksp PbSO4 6.3X 10 Answer= 6.38 10 5 m/ solution since Na₂ so is strong electrolyte. This means It will completely dissociate. Se we draw ICE table for Na2S04: Na₂SO4 - 2Na+ + 5o2- I: 0.0lom OM OM 0.02OM o.olom OM so Initial concentration of son ion is 0.010M in lead sulfate solution. since lead sulfate is weak electrolyte. so It will not dissociate completely. Let s is the concentration of lead (molar solubility) of PbSO4. So we draw ICE Table. PbSO4 = Pb2+ + so 2- I:- OM 0.010M C: - + SM + SM E SM (0.0L+AM

Now we write expression of Ksp for PbSO4 Ksp = [Pb2+ ] [502] - ( where Pb²+ = concentration of 504² = Concentration of Lead in solution sulfate in solution so ... [Pb2+] = s Ksp = A x (0.010+ s) 1 [502] = 0.0lots since value of s is very small. so 10.0L0 + s = 0.010) so o.olors= 0.010) До s <<< 0.010 Ksp = sxo.010 Ksp = 6.38 1072 6.3X10 7 = 0.010s s = 6.3x 10-7 0.010 = 6.3x 105 m so concentration of Lead solution - s = 6.38 20-SM in saturated
summary:

For solving this type of problem, you first know that Na2SO4 is strong electrolytes. So it will dissociate completely. This means the concentration of sulfate ion is 0.010M in saturated solution of PbSO4 containing 0.010 M Na2SO4 due to dissociation of 0.010M of Na2SO4.

So initial concentration of Sulfate in saturated solution of PbSO4 is 0.010M while initial concentration of lead in PbSO4 saturated solution is 0 M.

Since PbSO4 is weak electrolyte . So it will not dissociate completely. Let s be the molar solubility of PbSO4. So we write expression of Ksp for PbSO4 to calculate value of s.

Remember:

Molar solubility s is very small compared to 0.010M

So 0.010+s = 0.010M

So after calculation,

Answer = 6.3 × 10 ^-5 M