Question

A. Calculate the molar solubility of PbBr₂ in a 0.2890 M lead(II) nitrate, Pb(NO₃)₂ solution. The K_sp of lead(II) bromide is 6.60 ✕ 10⁻⁶.

B. Let's say we have a beaker where a saturated solution of lead(II) bromide is in equilibrium with solid lead(II) bromide. In which of these cases will the molar solubility be lowest after equilibrium is reestablished?

a. Upon the addition of more water.

b. After the addition of 0.180 moles of Br⁻ ion.

c. Not enough information given.

d. None of these things will do it.

e. After the addition of 0.180 moles of Pb²⁺ ion.

f. After the addition of solid NaNO₃.

g. After letting some of the solvent evaporate.

Answer 1

Pb Br2(s) --------------> Pb+2 (aq) + 2Br-(aq)

s 0.289 +s 2s due to common ion

since s is very small compared to 0.289 ,

s 0.289 2s

Thus ksp = 6.60x10^-6 = 0.289 x(2s)²

or s = 2.389x10^-3 M

part B Option B is correct choice

when [Br-] = 0.180 M

solubility =[Pb+2] = ksp /[Br-}²

   = 6.6x10 ^-6 / (0.180)²

= 2.03 x10^-4 M

this is the lowest solubility of PbBr2

By adding water or solid NaNO3(no common ion) solubility does onot get affected.

Whe 0.180M [Pb+2] is addded

solubility = [Br-] = {ksp/[pb+2]}^1/2

   ^=[ 6.60x10^-6 /0.180]^1/2

     = 6.05x10^-3 M

This is more than when 0.180M br- addded.