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.