Question

estimate the ksp of Lindocaine*HCl, soluble 1 part in 0.7 part water.

Answer 1

soluble 1 part in 0.7 part water means 1 g of Lindocaine*HCl dissolves in 0.7 g water.

Molar mass of Lindocaine*HCl = 270.84 g/mol

1 g of Lindocaine*HCl = 1g/(270.84 g/mol) = 0.00369 moles

0.7 g water = 0.0007kg = 0.0007 L [as density of water = 1 kg/L]

[Lindocaine*HCl] = 0.00369/0.0007 = 5.27 moles/L = 5.27 M

Lindocaine*HCl <---> Lindocaine.H⁺ +     Cl^-

5.27 M                                5.27M                  5.27M

Ksp = [Lindocaine.H⁺ ][Cl^-] = 5.27*5.27 = 27.78