Question

When \(200.0 \mathrm{mL}\) of \(0.350 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq})\) are added to 200.0 mL of 0.0100 M AgNO 3(aq), what percentage of the \(\mathrm{Ag}^{+}\) is left unprecipitated?

Short answer

The percentage of unprecipitated \(Ag^{+}\) is 0%.

Step-by-step solution

Identify the reaction

The first step is identifying the reaction. When \(AgNO_3\) and \(K_2CrO_4\) are mixed, they react to form \(Ag_2CrO_4\) and \(KNO_3\), with the \(Ag_2CrO_4\) precipitating out. The balanced chemical equation for this reaction is: \[2AgNO_3(aq) + K_2CrO_4(aq) \rightarrow Ag_2CrO_4(s) + 2KNO_3(aq)\]

Method to Calculate the Remaining Ag+

Calculate the initial number of moles of \(\mathrm{Ag}^{+}\) and \(K_2CrO_4\) respectively. The formula to calculate moles is given by \(Molarity = moles/volume\). Hence moles = Molarity x Volume. \[Moles_{Ag^{+}}= 0.0100 \, Molarity \times 200.0 \, mL = 0.002 \, moles\] \[Moles_{K_2CrO_4}= 0.350 \, Molarity \times 200.0 \, mL = 0.0700 \, moles\] Next, find out how much \(Ag^{+}\) will get precipitated out by \(K_2CrO_4\). From the equation, we can see that 1 mol of \(K_2CrO_4\) reacts with 2 mol of \(Ag^+\). So, \(0.0700 \, mol of \, K_2CrO_4\) should react with \(2 \times 0.0700 = 0.0140 \, mol of \, Ag^+\).

Calculate percentage of unreacted Ag+

Calculate the remaining \(Ag^{+}\) by subtracting reacted \(Ag^{+}\) from initial \(Ag^{+}\). \[Remaining \, Ag^{+} = Initial \, Ag^{+} - Reacted \, Ag^{+} = 0.002 \, mol - 0.0140 \, mol = -0.012 \, mol\] Since the \(Remaining \, Ag^{+}\) is negative, it means that all \(Ag^{+}\) has reacted and none remained unreacted thus leading to no unprecipitated \(Ag^{+}\). So, the percentage of unprecipitated \(Ag^{+}\) is 0%.

Key concepts

Molarity Calculations

Molarity is a key concept in chemistry that helps us understand how concentrated a solution is. It is defined as the number of moles of solute dissolved in one liter of solution. Calculating molarity involves two main components:

• Moles of Solute: The amount of substance, usually measured in moles, that is dissolved in the solution.
• Volume of Solution: The total volume of the solution, typically measured in liters.

Molarity can be calculated using the formula: \[ \text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}}\]To solve molarity calculations efficiently:

• First, ensure the volume is in liters before performing calculations.
• Use the formula to find the number of moles when molarity and volume are given, or determine molarity if moles and volume are known.
• For everyday problems, also remember that milliliters can be converted to liters by dividing by 1000.

In the provided exercise, we calculated the initial moles of both reactants using their molarity and given volumes. This step is critical for further analysis, like determining which reactant limits the reaction.

Precipitation Reaction

In chemistry, precipitation reactions are crucial because they result in the formation of a solid from a solution. If two solutions containing soluble salts are mixed, an insoluble salt (a precipitate) may form. The conditions of such reactions depend on:

• The solubility rules that determine whether a solid will form or not.
• The particular ions present in the reactants.

For the exercise, we observe a precipitation reaction between silver nitrate \((AgNO_3)\) and potassium chromate \((K_2CrO_4)\). The reaction is given by:\[ 2AgNO_3(aq) + K_2CrO_4(aq) \rightarrow Ag_2CrO_4(s) + 2KNO_3(aq)\]The formation of the silver chromate \((Ag_2CrO_4)\) solid is the precipitation in this reaction.To predict the occurrence of such a reaction, you can:

• Write out the balanced chemical equation.
• Determine whether any of the products are insoluble.

Understanding these reactions helps in devising methods for isolating or removing specific ions from solutions.

Stoichiometry

Stoichiometry is a powerful tool in chemistry, which involves the calculation of reactants and products in chemical reactions. It uses the principles of molar relationships in balanced chemical equations to determine the precise amounts necessary or produced.In solving stoichiometry problems:

• First, ensure you have a balanced chemical equation.
• Use mole ratios from the equation to calculate the amounts of reactants needed or products formed.
• Identify limiting reactants, which determine how far a reaction can proceed.

In the original exercise, stoichiometry was key in understanding the reaction between \(AgNO_3\) and \(K_2CrO_4\). We calculated that 0.0700 moles of \(K_2CrO_4\) could potentially react with 0.0140 moles of \(Ag^+\) ions, fully using these and leading to the formation of \(Ag_2CrO_4\). By recognizing that all \(Ag^+\) ions precipitate out, stoichiometry demonstrated that none of the silver ions remained in the solution, confirming a complete reaction.By applying stoichiometry, you can predict reaction outcomes and efficiently use reactants.