Question

A.) Calculate the solubility in g/L of AgBr in pure water.

B.) Calculate the solubility in g/L of AgBr in 0.0018 M NaBr. (Enter answer in scientific notation.)

Answer 1

Ksp of AgBr = 5.0*10^-13

molar mass of AgBr = 187.77 g/mol

A)

AgBr   ßà Ag+   + Br-

               s        s

Ksp = s^2

5.0*10^-13 = s^2

s = 7.071*10^-7 mol/L

   = 7.071*10^-7*187.77 g/L

    =1.335*10^-4 g/L

Answer: 1.335*10^-4 g/L

B)

Ksp = [Ag+] [Br-]

5.0*10^-13 = s*0.0018

s = 2.78 *10^-10 mol/L

   = 2.78*10^-10*187.77 g/L

    =5.22 *10^-8 g/L

Answer: 5.22 *10^-8 g/L