How do we calculate ionic strength on the molal scale?

It depends on how concentrated your solution is and what it contains. It becomes simplest for ideally-dilute solutions containing strong electrolytes wherein we ignore ion pairing.

If we do that, then the ionic strength in terms of molality is given as follows (Physical Chemistry, Levine, pg. 312):

#I_m = 1/2sum_i z_i^2m_i#

where for a strong electrolyte represented by

#M_(nu_(+))X_(nu_(-))(s) stackrel(H_2O(l)" ")(->) nu_(+)M^(z_(+))(aq) + nu_(-)X^(z_(-))(aq)#,

we have:

• #m_i# is the molality of the entire #i#th strong electrolyte.

For instance, a cation within #"HCl"# would have a molality of #m_+ = nu_+m_i#, and an anion within #"HCl"# would have a molality of #m_(-) = nu_(-)m_i#. The #m_i# would then correspond to #"HCl"#.

In other words, the stoichiometry of the ion gives its contribution factor.

• #z_i# is its charge.

For instance, a cation would have a charge #z_(+)# and anion would have a charge of #z_(-)#. This charge has both magnitude and sign, but the sign goes away by squaring.

EXAMPLE: HCl

Then for simplicity, first consider a solution containing only one strong electrolyte #"HCl"# of concentration #"0.01 m"#.

Its ionic strength is given by:

#I_m = 1/2 (z_(H^(+))^2m_(H^(+)) + z_(Cl^(-))^2m_(Cl^(-)))#

#= 1/2(z_(H^(+))^2nu_(H^(+))m_(HCl) + z_(Cl^(-))^2nu_(Cl^(-))m_(HCl))#

But since they are the same charge magnitudes, #|z_(+)z_(-)| = z_(pm)^2#.

#=> 1/2(|z_(H^(+))z_(Cl^(-))|nu_(H^(+))m_(HCl) + |z_(H^(+))z_(Cl^(-))|nu_(Cl^(-))m_(HCl))#

#= 1/2|z_(H^(+))z_(Cl^(-))|(nu_(H^(+))m_(HCl) + nu_(Cl^(-))m_(HCl))#

#= 1/2|z_(H^(+))z_(Cl^(-))|(nu_(H^(+)) + nu_(Cl^(-)))m_(HCl)#

And so,

#color(blue)(I_m) = 1/2|1 cdot -1| cdot (1 + 1) cdot "0.01 m" = color(blue)("0.01 mol solute/kg solvent")#

That should be no surprise. A 1:1 electrolyte should have the same molality in solution as it would before it dissociates.

EXAMPLE: THREE-ELECTROLYTE SOLUTION

Now let's say you had a more complicated solution. Let's say we had three strong electrolytes:

• #"0.01 m"# #"NaCl"#
• #"0.02 m"# #"HCl"#
• #"0.03 m"# #"BaCl"_2#

Again, ignoring ion pairing. We instead get for our initial ionic strength expression:

#I_m = 1/2sum_i z_i^2 m_i#

#= 1/2[z_(Na^(+))^2m_(Na^(+)) + z_(H^(+))^2m_(H^(+)) + z_(Ba^(2+))^2m_(Ba^(2+)) + z_(Cl^(-))^2m_(Cl^(-))]#

We should note that the chloride comes from all three electrolytes, so #nu_(Cl^(-)) = 4# in total:

#"NaCl"(aq) -> "Na"^(+)(aq) + "Cl"^(-)#
#"HCl"(aq) -> "H"^(+)(aq) + "Cl"^(-)(aq)#
#"BaCl"_2(aq) -> "Ba"^(2+)(aq) + 2"Cl"^(-)(aq)#

This means:

#I_m = 1/2[|z_(+)z_(-)|m_(Na^(+)) + |z_(+)z_(-)|m_(H^(+)) + z_(Ba^(2+))^2m_(Ba^(2+)) + |z_(+)z_(-)|m_(Cl^(-))]#

#= 1/2[|z_(+)z_(-)|nu_(Na^(+))m_(NaCl) + |z_(+)z_(-)|nu_(H^(+))m_(HCl) + z_(Ba^(2+))^2nu_(Ba^(2+))m_(BaCl_2) + |z_(+)z_(-)|(nu_(Cl^(-))m_(NaCl) + nu_(Cl^(-))m_(HCl) + nu_(Cl^(-))m_(BaCl_2))]#

From this we then get:

#color(blue)(I_m) = 1/2[|1 cdot -1|cdot 1 cdot "0.01 m NaCl" + |1 cdot -1|cdot 1 cdot "0.02 m HCl" + 2^2 cdot 1 cdot "0.03 m BaCl"_2 + |1 cdot -1|(1 cdot "0.01 m NaCl" + 1 cdot "0.02 m HCl" + 2 cdot "0.03 m BaCl"_2)]#

#=# #color(blue)("0.12 mol solutes/kg solvent")#

That's the rigorous way to do it. You could also just assign the concentration of the electrolyte to the ion, and multiply by its charge squared, then add it all up based on the concentration each electrolyte contributes.

#color(blue)(I_m) = 1/2[1^2 cdot 1 cdot "0.01 m Na"^(+) + {1^2 cdot 1 cdot "0.01 m Cl"^(-) + 1^2 cdot 1 cdot "0.02 m Cl"^(-) + 1^2 cdot 2 cdot "0.03 m Cl"^(-)} + 1^2 cdot 1 cdot "0.02 m H"^(+) + 2^2 cdot 1 cdot "0.03 m Ba"^(2+)]#

#=# #color(blue)("0.12 mol solutes/kg solvent")#