Question

3. What is the concentration of Pb if PbBris placed in a 0.05M solution of NaBr? (You may assume no complexes are formed.) 4. Write all necessary equations (Ka, Kb MB etc etc) needed to determine exactly the [H'] of a 0.05M solution of Ca(HCOO): (calcium formate). (Calcium formate is completely soluble in water and you can assume no Ca(OH)2 is formed. ) Ignore activites

PbBr2: 6.6*10^-6

Answer 1

Solution:

3) Solubility (s) can be calculated from Ksp.

PbBr2 = Pb2+ + 2Br-

Ksp = s x (2s)^2 = 4s^3

s^3 =Ksp / 4 = 6.6 x 10^-6 / 4 = 1.65 x 10^-6

s = 1.18 x 10^-2 M = 0.0118 M

Therefore,

[Pb2+] = s = 0.0118 M

Total concentration of Pb2+ in 0.05 M NaBr is same because no other products of Pb will forms.

Hence, concentration of Pb2+ =  0.0118 M

= 0.0618 M