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Ch.4 - Chemical Quantities & Aqueous Reactions
All textbooksTro 4th EditionCh.4 - Chemical Quantities & Aqueous ReactionsProblem 12
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Chapter 4, Problem 12

We prepare a solution by mixing 0.10 L of 0.12 M sodium chloride with 0.23 L of a 0.18 M MgCl2 solution. What volume of a 0.20 M silver nitrate solution is needed to precipitate all the Cl- ions in the solution as AgCl?

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1
Calculate the moles of Cl- ions from the NaCl solution using the formula: \( \text{moles} = \text{volume} \times \text{molarity} \).
Calculate the moles of Cl- ions from the MgCl2 solution. Remember that each mole of MgCl2 provides 2 moles of Cl- ions.
Add the moles of Cl- ions from both solutions to find the total moles of Cl- ions present.
Use the stoichiometry of the reaction between Cl- ions and Ag+ ions to determine the moles of Ag+ needed. The reaction is: \( \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \).
Calculate the volume of 0.20 M AgNO3 solution required to provide the necessary moles of Ag+ ions using the formula: \( \text{volume} = \frac{\text{moles}}{\text{molarity}} \).
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