1. Calculate the ionic strength of thefolowing 0.2 m aqueous solutions. I -IXvibjz where b, is the molality (b/bo) and v, is the number of ions assuming that the electrolyte is fully dissociated. (a) Na2Se (b) CdS (c) MgBr2 (d) FeCl3 (e) Na3PO4 (f) Al2(S0)3 (g) V(NO3)s Solution (a) Na 2 Se (aq) <======> 2 Na + (aq) + Se 2- (aq) We have v i (Na + ) = 2; v i (Se 2- ) = 1 b i (Na + ) = 2*0.2 m = 0.4 m; b i (Se 2- ) = 1*0.2 m = 0.2 m z i (Na + ) = +1; z i (Se 2- ) = -2 The ionic strength of the solution is I = ½*[(2)*(0.2 m)*(+1) 2 + (1)*(0.2 m)*(-2) 2 ] = ½*[0.4 m + 0.8 m] = ½*(1.2 m) = 0.6 m (ans). (b) CdS (aq) <======> Cd 2+ (aq) + S 2- (aq) We have v i (Cd 2+ ) = 1; v i (S 2- ) = 1 b i (Na + ) = 1*0.2 m = 0.2 m; b i (S 2- ) = 1*0.2 m = 0.2 m z i (Cd 2+ ) = +2; z i (S 2- ) = -2 The ionic strength of the solution is I = ½*[(1)*(0.2 m)*(+2) 2 + (1)*(0.2 m)*(-2) 2 ] = ½*[0.8 m + 0.8 m] = ½*(1.6 m) = 0.8 m (ans). (c) MgBr 2 (aq) <======> Mg 2+ (aq) + 2 Br - (aq) We have v i (Mg 2+ ) = 1; v i (Br - ) = 2 b i (Mg 2+ ) = 1*0.2 m = 0.2 m; b i (Br - ) = 2*0.2 m = 0.4 m z i (Mg 2+ ) = +1; z i (Br - ) = -1 The ionic strength of the solution is I = ½*[(1)*(0.2 m)*(+2) 2 + (2)*(0.4 m)*(-1) 2 ] = ½*[0.8 m + 0.8 m] = ½*(1.6 m) = 0.8 m (ans). (d) FeCl 3 (aq) <======> Fe 3+ (aq) + 3 Cl - (aq) We have v i (Fe 3+ ) = 1; v i (Cl - ) = 3 b i (Fe 3+ ) = 1*0.2 m = 0.2 m; b i (Cl - ) = 3*0.2 m = 0.6 m z i (Fe 3+ ) = +3; z i (Cl - ) = -1 The ionic strength of the solution is I = ½*[(1)*(0.2 m)*(+3) 2 + (3)*(0.6 m)*(-1) 2 ] = ½*[1.8 m + 1.8 m] = ½*(3.6 m) = 1.8 m (ans). .

1. 1. Calculate the ionic strength of thefolowing 0.2 m aqueous solutions. I -IXvibjz where b, is the
molality (b/bo) and v, is the number of ions assuming that the electrolyte is fully dissociated. (a)
Na2Se (b) CdS (c) MgBr2 (d) FeCl3 (e) Na3PO4 (f) Al2(S0)3 (g) V(NO3)s
Solution
(a) Na 2 Se (aq) <======> 2 Na +
(aq) + Se 2-
(aq)
We have
v i (Na +
) = 2; v i (Se 2-
) = 1
b i (Na +
) = 2*0.2 m = 0.4 m; b i (Se 2-
) = 1*0.2 m = 0.2 m
z i (Na +
) = +1; z i (Se 2-
) = -2
The ionic strength of the solution is
I = ½*[(2)*(0.2 m)*(+1) 2
+ (1)*(0.2 m)*(-2) 2
] = ½*[0.4 m + 0.8 m] = ½*(1.2 m) = 0.6 m
(ans).
(b) CdS (aq) <======> Cd 2+
(aq) + S 2-
(aq)
We have
v i (Cd 2+
) = 1; v i (S 2-
) = 1
b i (Na +
) = 1*0.2 m = 0.2 m; b i (S 2-
) = 1*0.2 m = 0.2 m
z i (Cd 2+
) = +2; z i (S 2-
) = -2
The ionic strength of the solution is
I = ½*[(1)*(0.2 m)*(+2) 2
+ (1)*(0.2 m)*(-2) 2
] = ½*[0.8 m + 0.8 m] = ½*(1.6 m) = 0.8 m
(ans).

2. (c) MgBr 2 (aq) <======> Mg 2+
(aq) + 2 Br -
(aq)
We have
v i (Mg 2+
) = 1; v i (Br -
) = 2
b i (Mg 2+
) = 1*0.2 m = 0.2 m; b i (Br -
) = 2*0.2 m = 0.4 m
z i (Mg 2+
) = +1; z i (Br -
) = -1
The ionic strength of the solution is
I = ½*[(1)*(0.2 m)*(+2) 2
+ (2)*(0.4 m)*(-1) 2
] = ½*[0.8 m + 0.8 m] = ½*(1.6 m) = 0.8 m
(ans).
(d) FeCl 3 (aq) <======> Fe 3+
(aq) + 3 Cl -
(aq)
We have
v i (Fe 3+
) = 1; v i (Cl -
) = 3
b i (Fe 3+
) = 1*0.2 m = 0.2 m; b i (Cl -
) = 3*0.2 m = 0.6 m
z i (Fe 3+
) = +3; z i (Cl -
) = -1
The ionic strength of the solution is
I = ½*[(1)*(0.2 m)*(+3) 2
+ (3)*(0.6 m)*(-1) 2
] = ½*[1.8 m + 1.8 m] = ½*(3.6 m) = 1.8 m
(ans).