Browse: Home / Uncategorized / What volume of a 0.21 M silver nitrate solution is required to precipitate all the Cl- ion in the solution as?
By admin on December 10, 2011
A solution is prepared by mixing 0.11 L of 0.14 M sodium chloride with 0.20 L of a 0.19 M MgCl2 solution.
-What volume of a 0.21 M silver nitrate solution is required to precipitate all the Cl- ion in the solution as AgCl?
Posted in Uncategorized | Tagged Cl, ion, mgcl2, silver, silver nitrate solution, sodium chloride |
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Total moles of Cl- = 0.11 L X 0.14 mol/L + 2 X 0.20 L X 0.19 mol/L = 0.0914 mol Cl-
You will need as many moles of AgNO3 as you have moles of Cl-, so:
0.0914 mol / 0.21 mol/L =0.435 L AgNO3