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

Answer

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Answer

$I = 0.05 mol \cdot {kg}^{- 1}$

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

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

Step 3 ：Calculate the ionic strength (I) using the formula: $I = \frac{1}{2} \sum (c i \cdot z i^{2})$ , 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 ： $I = 0.05 mol \cdot {kg}^{- 1}$

13. A solution was prepared by mixing equal volume of 0.10 mol⋅kg−1NH4Cl and 0.10 mol⋅kg− 1NH3⋅H2O, its ionic strength ( I) is ( ) mol⋅kg−1.A. 0.05B. 0.075C. 0.10D. 0.15

Answer

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