45.8 mL of a 3.14 M sodium chloride solution were used to react completely with 50.0 mL of an aqueous silver nitrate solution.What is the molarity of the silver nitrate solution?

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The reaction between sodium chloride (NaCl) and silver nitrate (AgNO3) is a double displacement reaction, where NaCl reacts with AgNO3 to form silver chloride (AgCl) and sodium nitrate (NaNO3). The balanced chemical equation for this reaction is:$$ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 $$From the equation, we see that the mole ratio of NaCl to AgNO3 is 1:1. Therefore, the moles of NaCl will be equal to the moles of AgNO3 that reacted.First, calculate the moles of NaCl:Molarity (M) = moles of solute / volume of solution (L)Moles of NaCl = Molarity of NaCl solution * Volume of NaCl solution in liters$$ = 3.14 \, \text{M} \times 45.8 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} $$$$ = 0.143772 \, \text{moles} $$Since the mole ratio of NaCl to AgNO3 is 1:1, the moles of AgNO3 will also be 0.143772.Next, calculate the molarity of the AgNO3 solution:$$ \text{Molarity of AgNO}_3 = \frac{\text{Moles of AgNO}_3}{\text{Volume of AgNO}_3 \text{ solution in liters}} $$$$ = \frac{0.143772}{50.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}} $$$$ = 2.88 \, \text{M} $$Therefore, the molarity of the silver nitrate solution is 2.88 M.

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