Question

Calculate the ionic strength of (a) 0.008 7 M KOH and (b) 0.000 2 $\mathbf{M}\mathbf{-}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{La}{\mathbf{\left(}{\mathbf{IO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}$(assuming complete disassociation at this low concentration and no hydrolysis reaction to make${\mathbf{LaOH}}^{\mathbf{2}\mathbf{+}}$ ).

Short answer

The ionic strength for 0.008 7 M KOH is 0.0087M and for

0.000 2 $\mathrm{M}-\mathrm{La}{\left({\mathrm{IO}}_3\right)}_3$is 0.0012M

Step-by-step solution

Step 1:Finding the ionic strength of  0.008  7   M   KOH

The ionic strength $\mathrm{\mu }‐\mathrm{a}$a concentration of ions in a solution is calculated by using the following formula,

$\mathbf{\mu }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(c_1{z_1}^2+c_2{z_2}^2+...\right)$

, wherestands for concentration and z stands for charge.

First the ionic strength of 0.008 7 M KOH can be calculated.

The reaction of disassociation would be$\mathrm{KOH}\mathrm{֏}{\mathrm{k}}^{+}+{\mathrm{OH}}^{-}$

role="math" localid="1654835701912" $$\begin{gathered} {\mathrm{z}}_1=+1 \\ {\mathrm{z}}_2=-1 \\ \mathrm{\mu }=\frac{1}{2}\left(0.0087\times {\left(+1\right)}^2+0.0087\times {\left(-1\right)}^2\right)-0.0087\mathrm{M} \end{gathered}$$

Finding the ionic strength of 0.000   2 M-    La(IO3)3

Next, for 0.000 2 $\mathrm{M}-\mathrm{La}{\left({\mathrm{IO}}_3\right)}_3$, in this case both ions have different concentration.

$$\begin{gathered} \mathrm{La}{\left({\mathrm{IO}}_3\right)}_3+{\mathrm{H}}_3\mathrm{O֏}\mathrm{La}{\left(\mathrm{OH}\right)}^2+3{\mathrm{IO}}_3+{\mathrm{H}}^{+} \\ \mathrm{\mu }=\frac{1}{2}\left(0.0002\times {\left(+3\right)}^2+0.0006\times {\left(-1\right)}^2\right)=0.0012\mathrm{M} \end{gathered}$$

Thus, the ionic strength for 0.008 7 M KOH is 0.0087M and for $0.0002\mathrm{M}-\mathrm{La}{\left({\mathrm{IO}}_3\right)}_3$is 0.0012 M