If 25.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.842 g of precipitate, what is the molarity of silver ion in the original solution?

To find the molarity of the silver ion (\( \text{Ag}^+ \)) in the original solution, we can break the problem down into the following steps:

1. **Identify the reaction:**  
   The reaction between silver nitrate (\( \text{AgNO}_3 \)) and potassium chloride (\( \text{KCl} \)) produces silver chloride (\( \text{AgCl} \)) precipitate and potassium nitrate (\( \text{KNO}_3 \)) as a by-product:
   \[
   \text{AgNO}_3(aq) + \text{KCl}(aq) \rightarrow \text{AgCl}(s) + \text{KNO}_3(aq)
   \]
   From this, we see that 1 mole of \( \text{AgNO}_3 \) reacts with 1 mole of \( \text{KCl} \) to produce 1 mole of \( \text{AgCl} \).

2. **Calculate the moles of \( \text{AgCl} \) formed:**  
   The molar mass of silver chloride (\( \text{AgCl} \)) is the sum of the atomic masses of silver (\( 107.87 \, \text{g/mol} \)) and chlorine (\( 35.45 \, \text{g/mol} \)):
   \[
   \text{Molar mass of } \text{AgCl} = 107.87 + 35.45 = 143.32 \, \text{g/mol}
   \]
   Now, using the mass of the precipitate (0.842 g), we can calculate the number of moles of \( \text{AgCl} \):
   \[
   \text{Moles of } \text{AgCl} = \frac{0.842 \, \text{g}}{143.32 \, \text{g/mol}} = 0.005876 \, \text{mol}
   \]

3. **Determine the moles of \( \text{Ag}^+ \):**  
   Since 1 mole of \( \text{AgCl} \) is produced by 1 mole of \( \text{Ag}^+ \), the moles of \( \text{AgCl} \) is equal to the moles of \( \text{Ag}^+ \). Therefore, the moles of \( \text{Ag}^+ \) is:
   \[
   \text{Moles of } \text{Ag}^+ = 0.005876 \, \text{mol}
   \]

4. **Calculate the molarity of \( \text{Ag}^+ \):**  
   Molarity is given by the formula:
   \[
   \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}
   \]
   The volume of the solution is 25.0 mL, which is equal to 0.0250 L. Thus, the molarity of \( \text{Ag}^+ \) is:
   \[
   \text{Molarity of } \text{Ag}^+ = \frac{0.005876 \, \text{mol}}{0.0250 \, \text{L}} = 0.235 \, \text{M}
   \]

Thus, the molarity of \( \text{Ag}^+ \) in the original solution is **0.235 M**.

Would you like any further details or explanations?

Here are 5 related questions:
1. How would you calculate the molarity if the volume of the solution was different?
2. What if the precipitate was a different compound, like \( \text{PbCl}_2 \)?
3. How would the calculation change if some \( \text{AgCl} \) remained dissolved in the solution?
4. Can the molarity of \( \text{K}^+ \) in the solution also be calculated from this reaction?
5. How could you experimentally determine the mass of the \( \text{AgCl} \) precipitate more accurately?

**Tip:** Always ensure that the units of volume are converted to liters when calculating molarity, as the standard unit for molarity is mol/L.

Learn how to calculate the molarity of silver ion (Ag+) in a silver nitrate solution when 0.842 g of AgCl precipitate is formed from 25.0 mL of solution. This problem covers stoichiometry, molarity calculations, and chemical reactions, suitable for Grades 10-12.

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