Question

Find the ionic strength

a) A solution of 0.00499 M FeSO3

b) A solution of 0.00128 M CuCl2

c) A solution of 0.000736 M HCl and 0.000474 M La(NO3)3

d) 0.00040 M La(IO3)3 Assume complete dissociation at this low concentration and no hydrolysis reaction to form LaOH2 .

Please answer all parts and show answer

Answer 1

a)
FeSO3 ---->Fe2+ + SO3-
ionic strength = (1/2) sum of [ci*zi^2]
ci is concentration of ion
zi is charge on ion

ionic strength = (1/2) [0.00499*(+2)^2 + 0.00499*(-2)^2]
= 0.01996

b)
CuCl2 ---> Cu2+ + CL-
[Cu2+ ] = 0.00128 M
[Cl-] = 2*0.00128 M = 0.00256 M
ionic strength = (1/2) [0.00128*(+2)^2 + 0.00256*(-1)^2]
= 0.00384

C)
[H+] = 0.000736 M
[Cl-] = 0.000736 M
[La3+] = 0.000474 M
[NO3-] = 3*0.000474 M = 0.001422 M

ionic strength = 0.5* [0.000736*(1)^2 + 0.000736*(-1)^2 + 0.000474*(+3)^2 + 0.001422*(-1)^2]
= 0.00358

d)
[La3+] = 0.00040 M
[IO3-] = 3*0.00040 = 0.0012 M
ionic strength = 0.5* [0.00040*(+3)^2 + 0.00012*(-1)^2]
=0.0024