13. A solution was prepared by mixing equal volume of $0.10 \mathrm{~mol} \cdot \mathrm{kg}^{-1} \mathrm{NH}_{4} \mathrm{Cl}$ and $0.10 \mathrm{~mol} \cdot \mathrm{kg}^{-}$ ${ }^{1} \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$, its ionic strength ( I) is ( ) $\mathrm{mol} \cdot \mathrm{kg}^{-1}$. A. 0.05 B. 0.075 C. 0.10 D. 0.15

Step 1 ：Given equal volumes of $0.10 \mathrm{~mol} \cdot \mathrm{kg}^{-1} \mathrm{NH}_{4} \mathrm{Cl}$ and $0.10 \mathrm{~mol} \cdot \mathrm{kg}^{-1} \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ solutions are mixed.

Step 2 ：The final concentration of NH4+ and Cl- ions will be half of the initial concentration: $0.05 \mathrm{~mol} \cdot \mathrm{kg}^{-1}$ each.

Step 3 ：Calculate the ionic strength (I) using the formula: $I = \frac{1}{2} \sum(ci \cdot zi^2)$, where ci is the concentration of the ith ion and zi is the charge of the ith ion.

Step 4 ：Considering only NH4+ and Cl- ions: $I = \frac{1}{2} (0.05 \cdot 1^2 + 0.05 \cdot (-1)^2)$

Step 5 ：\(I = \frac{1}{2} (0.05 + 0.05)\)

Step 6 ：\(\boxed{I = 0.05 \mathrm{~mol} \cdot \mathrm{kg}^{-1}}\)