Calculate the ionic strength of the 2:2 electrolyte FeSO4 at a concentration [latex]c=0.01~\mathrm{mol~dm}^{-3}[/latex].

Inserting charges in Equation (7.31): [latex]I=\frac{1}{2}\{[{\bf F e}^{2+}]\times(+2)^{2}+[{ S O}_{4}^{2-}]\times(-2)^{2}\}[/latex] [latex]I=\frac{1}{2}\sum_{i=1}^{i=i}c_{i}z_{i}^{2}[/latex] (7.31) We next insert concentrations, again noting that one ferrous ion and one sulphate ion are formed per formula unit: [latex]{I}=\frac{1}{2}\{([\mathrm{FeSO}_{4}]\times4)+([\mathrm{FeSO}_{4}]\times4)\}[/latex] so we obtain the result I = 4 × c for this, a 2:2 electrolyte. Inserting the concentration c of [latex]\mathrm{[FeSO_{4}]}=0.01\ \mathrm{mol}\ \mathrm{dm}^{-3}[/latex], we obtain I = 0.04 mol [latex]\mathrm{dm}^{-3}[/latex], which explains why hard water containing [latex]\mathrm{FeSO}_{4}[/latex] has a greater influence than table salt of the same concentration.

Question 7.13: Calculate the ionic strength of the 2:2 electrolyte FeSO4 at......

Physical Chemistry: Understanding our Chemical World [1013861]

Calculate the ionic strength of the 2:2 electrolyte FeSO4 at a concentration $c=0.01~\mathrm{mol~dm}^{-3}$ .

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Inserting charges in Equation (7.31):

$I=\frac{1}{2}\{[{\bf F e}^{2+}]\times(+2)^{2}+[{ S O}_{4}^{2-}]\times(-2)^{2}\}$

$I=\frac{1}{2}\sum_{i=1}^{i=i}c_{i}z_{i}^{2}$ (7.31)

We next insert concentrations, again noting that one ferrous ion and one sulphate ion are formed per formula unit:

${I}=\frac{1}{2}\{([\mathrm{FeSO}_{4}]\times4)+([\mathrm{FeSO}_{4}]\times4)\}$

so we obtain the result I = 4 × c for this, a 2:2 electrolyte.
Inserting the concentration c of $\mathrm{[FeSO_{4}]}=0.01\ \mathrm{mol}\ \mathrm{dm}^{-3}$ , we obtain I = 0.04 mol $\mathrm{dm}^{-3}$ , which explains why hard water containing $\mathrm{FeSO}_{4}$ has a greater influence than table salt of the same concentration.