Question

How many grams of \(\mathrm{NaCl}\) are required to precipitate most of the Ag ions from \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.0113 \mathrm{M} \mathrm{AgNO}_{3}\) solution? Write the net ionic equation for the reaction.

Short answer

0.165 grams of NaCl are needed. Net ionic equation: \( \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \).

Step-by-step solution

Understand the Reaction

The reaction between sodium chloride (NaCl) and silver nitrate (AgNO₃) will result in the formation of silver chloride (AgCl) precipitate and sodium nitrate (NaNO₃). The net ionic equation of this reaction is: \[ \text{Ag}^+ (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) \]

Convert Volume to Liters

The volume of the AgNO₃ solution is given as 250 mL. Convert this to liters by dividing by 1000: \[ 250\, \text{mL} = 0.250 \, \text{L} \]

Calculate Moles of Ag⁺ Ions

Use the molarity equation \( \text{moles} = \text{molarity} \times \text{volume} \) to find the moles of \( \text{Ag}^+ \) ions. \[ \text{moles of Ag}^+ = 0.0113 \, \text{M} \times 0.250\, \text{L} = 0.002825 \text{ mol} \]

Determine Moles of NaCl Needed

According to the net ionic equation, 1 mole of NaCl provides 1 mole of Cl⁻ ion, which reacts with 1 mole of Ag⁺ ion. Therefore, you need 0.002825 moles of NaCl to react with 0.002825 moles of Ag⁺ ions.

Calculate Mass of NaCl Required

Find the molar mass of NaCl, which is 58.44 g/mol. Use the moles obtained in Step 4 to find the grams of NaCl needed: \[ \text{mass} = \text{moles} \times \text{molar mass} \] \[ = 0.002825 \, \text{mol} \times 58.44 \, \text{g/mol} = 0.16498 \, \text{g} \]

Round Off the Answer

Finally, round the answer to three significant figures since the least number of significant figures in the problem is three: 0.165 g.

Key concepts

Net Ionic Equation

A net ionic equation simplifies a chemical reaction by removing the ions that do not participate directly in the formation of the precipitate. In the context of this exercise, when sodium chloride (NaCl) and silver nitrate (AgNO₃) react, they produce a silver chloride (AgCl) precipitate. This precipitate is important because it shows us the primary participants in the reaction.
The full ionic equation would show all ions in solution, but for simplicity, the net ionic equation focuses on just those involved in forming the precipitate. For our exercise, the net ionic equation becomes:

• \[ \text{Ag}^+ (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) \]

This equation highlights only the silver (\( \text{Ag}^+ \)) and chloride (\( \text{Cl}^- \)) ions that combine to form solid AgCl. Knowing how to write net ionic equations is crucial, as it helps in understanding the essence of the chemical reaction without overloading on less relevant details.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions mix, resulting in the formation of an insoluble solid called a precipitate. This is a common event in chemistry that often helps in separating or identifying particular substances.
In our exercise, when \( \text{NaCl} \) solution is mixed with \( \text{AgNO}_3 \), they react to form a cloudy precipitate of silver chloride (\( \text{AgCl} \)). This happens because \( \text{AgCl} \) is less soluble in water, so it appears as a solid:

• Initial compounds: Silver nitrate (\( \text{AgNO}_3 \)) and sodium chloride (\( \text{NaCl} \))
• Precipitate formed: Silver chloride (\( \text{AgCl} \))

Precipitation reactions are valuable tools in chemistry for isolating certain ions or compounds, as these reactions can be employed in diverse applications like purifying water or analyzing specific chemicals.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution (mol/L) and is denoted by the symbol \( M \). Understanding molarity is essential for calculating how much of a substance is needed for a specific solution requirement.
In this exercise, the silver nitrate (\( \text{AgNO}_3 \)) solution has a molarity of 0.0113 M, meaning there are 0.0113 moles of \( \text{AgNO}_3 \) per liter of the solution. We use this value to calculate the number of moles of \( \text{Ag}^+ \) ions present, which is necessary for determining the amount of NaCl required to precipitate silver ions.
Here's a quick peek at the formula for molarity: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]By multiplying the molarity by the volume of the solution (converted to liters), you can calculate the moles of solute.