Page 350 - Physical chemistry understanding our chemical world

ACTIVITY 317

We next insert concentration terms, noting that one sodium ion and
NaCl is called a ‘1:1
one chloride are formed per formula unit of sodium chloride (which
electrolyte’ because the
is why we call it a 1:1 electrolyte). Accordingly, the concentrations
formula unit contains
+
−
of the two ions, [Na ]and [Cl ], are the same as [NaCl], so
one anion and one
cation.
1
I = {([NaCl] × 1) + ([NaCl] × 1)}
2
so we obtain the result for a 1:1 electrolyte that I (NaCl) = [NaCl].
We obtain the result
Note that I has the same units as concentration: inserting I = c only for 1:1 (uni-
−3
the NaCl concentration [NaCl] = 0.01 mol dm , we obtain I = valent) electrolytes.
−3
0.01 mol dm .
Worked Example 7.13 Calculate the ionic strength of the 2:2 electrolyte FeSO 4 at a
−3
concentration c = 0.01 mol dm .
Inserting charges in Equation (7.31):
1 2 2
2+
2−
I = {[Fe ] × (+2) + [SO ] × (−2) }
4
2
We next insert concentrations, again noting that one ferrous ion and one sulphate ion are
formed per formula unit:
1
I = {([FeSO 4 ] × 4) + ([FeSO 4 ] × 4)}
2
so we obtain the result I = 4 × c for this, a 2:2 electrolyte.
−3
Inserting the concentration c of [FeSO 4 ] = 0.01 mol dm , we obtain I = 0.04 mol
−3
dm , which explains why hard water containing FeSO 4 has a greater influence than
table salt of the same concentration.

Worked Example 7.14 Calculate the ionic strength of the 1:2 electrolyte CuCl 2 ,again
−3
of concentration 0.01 mol dm .
Inserting charges into Equation (7.31):

1
2
2
−
2+
I = {[Cu ] × (+2) + [Cl ] × (−1) }
2
We next insert concentrations. In this case, there are two chloride Note how the calcu-
−
ions formed per formula unit of salt, so [Cl ] = 2 × [CuCl 2 ], but only lation requires the
one copper, so [Cu ] = [CuCl 2 ]. charge per anion,
2+
rather than the total
1
I = {([CuCl 2 ] × 4) + (2[CuCl 2 ] × 1)} anionic charge.
2
so we obtain the result I = 3 × c for this, a 1:2 electrolyte. And, I = 0.03 mol dm −3
−3
because [CuCl 2 ] = 0.01 mol dm .