Calculate the ionic strength of the 1:2 electrolyte [latex]\mathrm{CuCl}_{2}[/latex], again of concentration [latex]0.01~{\mathrm{mol~dm}}^{-3}[/latex].

Inserting charges in Equation (7.31): [latex]I=\frac{1}{2}\{[C{ u}^{2+}]\times(+2)^{2}+[C{ l}^{-}]\times(-1)^{2}\}[/latex] [latex]I=\frac{1}{2}\sum_{i=1}^{i=i}c_{i}z_{i}^{2}[/latex] (7.31) We next insert concentrations. In this case, there are two chloride ions formed per formula unit of salt, so [latex][Cl^{-}]=2\times[\mathrm{CuCl}_{2}][/latex], but only one copper, so [latex][Cu^{2+}]=[\mathrm{CuCl}_{2}][/latex]. [latex]I=\frac{1}{2}\{([\mathrm{CuCl}_{2}]\times4)+(2[\mathrm{CuCl}_{2}]\times1)\}[/latex] so we obtain the result I = 3 × c for this, a 1:2 electrolyte. And, [latex]I=0.03\ \mathrm{mol~dm}^{-3}[/latex] because [latex][{ C u C l}_{2}]=0.01~\mathrm{mol~dm}^{-3}[/latex].

Question 7.14: Calculate the ionic strength of the 1:2 electrolyte CuCl2, a......

Physical Chemistry: Understanding our Chemical World [1013861]

Calculate the ionic strength of the 1:2 electrolyte $\mathrm{CuCl}_{2}$ , again of concentration $0.01~{\mathrm{mol~dm}}^{-3}$ .

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Inserting charges in Equation (7.31):

$I=\frac{1}{2}\{[C{ u}^{2+}]\times(+2)^{2}+[C{ l}^{-}]\times(-1)^{2}\}$

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

We next insert concentrations. In this case, there are two chloride ions formed per formula unit of salt, so $[Cl^{-}]=2\times[\mathrm{CuCl}_{2}]$ , but only one copper, so $[Cu^{2+}]=[\mathrm{CuCl}_{2}]$ .

$I=\frac{1}{2}\{([\mathrm{CuCl}_{2}]\times4)+(2[\mathrm{CuCl}_{2}]\times1)\}$

so we obtain the result I = 3 × c for this, a 1:2 electrolyte. And, $I=0.03\ \mathrm{mol~dm}^{-3}$ because $[{ C u C l}_{2}]=0.01~\mathrm{mol~dm}^{-3}$ .