Question

Question #2 determine the ionic strength of each of the following solutions (a) 0.05 M MgCl2 ANS= (b) 0.05 M NaCI ANS= (c) 0.05 M NaOH plus 0.05 M MgCl2 (all in the same solution) ANS= (d) 0.05 M K2SO4 ANS= (e) water that is pH 2 from addition of HCI ANS=-

Answer 1

The formula for the ionic strength is

ΙΣci ex Zh

Here, I is the ionic strength, $c_i$ is the ion concentration and $Z_i$ is the charge on the ion.

(a)

0.05 M
MgCl2
solution contains
Mg2+
ions and $Cl^-$ ions.   

[Mg2+] = [MgCl2] Mg2+] = 0.05 M [CT"]= 2 x [MgCl2] C1 ] = 0.10 M

The charge on
Mg2+
ions and $Cl^-$ ions is +2 and -1 respectively.   

Substitue values in the formula for the ionic strength.

1 = 5x [(0.05 x (2)2) + (0.10 x (192)] 1 = $x (0.20 +0.10] I = 0.15

Hence, 0.05 M solution is 0.15.

MgCl2

(b)

0.05 M
Naci
solution contains $Na^+$ ions and $Cl^-$ ions.   

[Nat] = [NaCl] [Na+] = 0.05 M [CI] = [NaCl C1"]=0.05 M

The charge on $Na^+$ ions and $Cl^-$ ions is +1 and -1 respectively.   

Substitue values in the formula for the ionic strength.

1 - Σε κά? I = 3x [(0.05 x (1)?) + (0.05 x (12) 5x [0.05 + 0.05] I = 0,05

Hence, 0.05 M solution is 0.15.

Naci

(c)

0.05M
NaOH
plus 0.05 M
MgCl2
solution contains $Na^+$ ions, $OH^-$ ions,
Mg2+
ions and $Cl^-$ ions.   

[Na+] = [NaOH) Nat] = 0.05 M [OH-] = [NaOH [OH-] = 0.05 M

[Mg2+] = [MgCl2] Mg2+] = 0.05 M [CT"]= 2 x [MgCl2] C1 ] = 0.10 M

The charge on $Na^+$ ions and $OH^-$ ions is +1 and -1 respectively.   

The charge on
Mg2+
ions and $Cl^-$ ions is +2 and -1 respectively.   

Substitue values in the formula for the ionic strength.

OG'0 = 1 [Ot 0 + 070 + 90 0 + 0 0 * =I [(z(1) 01'0) + (z(7) x co'o) (z(1) co'o) + (z(1) * co'o)] *= 1 2 x 7 =

Hence, 0.05M
NaOH
plus 0.05 M
MgCl2
solution is 0.20.

(d)

0.05 M
K2SO4
solution contains $K^+$ ions and $SO_4^{2-}$ ions.   

[K+] = 2 x [K2S04 [K+] = 0.10 M [502] = [K2S04] 502-1 = 0.05 M

The charge on $K^+$ ions and $SO_4^{2-}$ ions is +1 and -2 respectively.   

Substitue values in the formula for the ionic strength.

1 - Σακ Ζι I = 3x [(0.10 x (102) + (0.05 x (232) 1-1 x [0,10 + 0.20] I = 0.15

Hence, 0.05 M solution is 0.15.

K2SO4