Problem: A solution is prepared by mixing 0.12 L of 0.13 M sodium chloride with 0.24 L of a 0.18 M MgCl2 solution.What volume of a 0.22 M silver nitrate solution do we need to precipitate all the Cl− ion in the solution as AgCI?

FREE Expert Solution

Step 1. total mol

NaCl → Na+ + Cl

0.12 L×0.13 mol NaClL×1 mol Cl-1 mol NaCl=0.0156 mol Cl

MgCl2 → Mg2+ + 2 Cl

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Problem Details

A solution is prepared by mixing 0.12 L of 0.13 M sodium chloride with 0.24 L of a 0.18 M MgCl2 solution.

What volume of a 0.22 M silver nitrate solution do we need to precipitate all the Cl− ion in the solution as AgCI?

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