Answer to Question #201623 in General Chemistry for Mickayla

Question #201623

The solubility of lead(II) fluoride in water is 0.0175% (w/v) at 29°C. Calculate the Ksp of this salt at this temperature

  1. 1.46 x 10-12 M
  2. 1.46 x 10-9 M
  3. 3.64 x 10-10 M
  4. 1.31 x 10-9 M
  5. 1.31 x 10-12 M
  6. 3.28 x 10-10 M
1
Expert's answer
2021-06-07T03:23:37-0400

Solution

Solubility PbF2=0.0175w/v

Molar mass = 245.2g/mol

PbF2(s)↔️ Pb2+(aq)+2F-(aq)

ksp=[Pb2+] [F-]2


Molar solubility=


Dissociation equation shows for every mole of PbF2 that dissociates, 1mol of Pb2+ and 2 mol of F- are produced


Pb2+= 7.137

F-=

Ksp=(7.137 2


=1.4533

=1.46 M





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