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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?

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One response to “What volume of a 0.21 M silver nitrate solution is required to precipitate all the Cl- ion in the solution as?”

  1. hcbiochem
    hcbiochem
    December 10, 2011 at 4:08 pm |

    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

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