Question

Solutions containing \(50.0 \mathrm{~mL}\) of \(1.0 \times 10^{-4} M \mathrm{AgNO}_{3}\) and 100. \(\mathrm{mL}\) of \(1.0 \times 10^{-4} \mathrm{M} \mathrm{NaCl}\) are mixed. Show by calculation whether or not a precipitate will form. Assume the volumes are additive. \(\left(K_{\mathrm{sp}}\right.\) for \(\left.\mathrm{AgCl}=1.7 \times 10^{-10}\right)\)

Short answer

A precipitate will form because \(Q = 2.22 \times 10^{-9}\) is greater than \(K_{sp}\).

Step-by-step solution

Calculate Initial Moles

First, calculate the moles of \(\text{AgNO}_3\) and \(\text{NaCl}\) using the formula \(n = C \times V\). For \(\text{AgNO}_3\:\) \[ n_{\text{AgNO}_3} = 1.0 \times 10^{-4} M \times 0.050 L = 5.0 \times 10^{-6} \text{ moles} \]For \(\text{NaCl}\:\) \[ n_{\text{NaCl}} = 1.0 \times 10^{-4} M \times 0.100 L = 1.0 \times 10^{-5} \text{ moles} \]

Calculate Final Volumes

Add the volumes of \(\text{AgNO}_3\) and \(\text{NaCl}\): \[ V_{\text{total}} = 50.0 \text{ mL} + 100.0 \text{ mL} = 150.0 \text{ mL} = 0.150 \text{ L} \]

Calculate Final Concentrations

Determine the final concentrations of \(\text{Ag}^+\) and \(\text{Cl}^-\) in the mixed solution: \[ [\text{Ag}^+] = \frac{5.0 \times 10^{-6} \text{ moles}}{0.150 \text{ L}} = 3.33 \times 10^{-5} M \]\[ [\text{Cl}^-] = \frac{1.0 \times 10^{-5} \text{ moles}}{0.150 \text{ L}} = 6.67 \times 10^{-5} M \]

Calculate the Reaction Quotient Q

Calculate the reaction quotient \(Q\) to determine if a precipitate will form: \[ Q = [\text{Ag}^+][\text{Cl}^-] \]\[ Q = (3.33 \times 10^{-5} M)(6.67 \times 10^{-5} M) = 2.22 \times 10^{-9} \]

Compare Q with K_{sp}

Compare the value of \(Q\) with \(K_{sp}\) for \(\text{AgCl}\). \[ K_{sp}(\text{AgCl}) = 1.7 \times 10^{-10} \]Since \(Q = 2.22 \times 10^{-9} \) and \(Q > K_{sp} \), a precipitate of \(\text{AgCl}\) will form.

Key concepts

Mole Calculations

Mole calculations are essential for understanding chemical reactions. Moles represent the amount of substance and can be calculated with the formula:

\[ n = C \times V \]

where \( n \) is the number of moles, \( C \) is the concentration in moles per liter (M), and \( V \) is the volume in liters (L).

In the provided problem, we calculated the moles of both \( \text{AgNO}_3 \) and \( \text{NaCl} \) before mixing them:

• \(\text{AgNO}_3 \)
• Calculated moles: \(\text{5.0} \times 10^{-6} \) moles


• \(\text{NaCl} \)
• Calculated moles: \(\text{1.0} \times 10^{-5} \) moles

Properly calculating moles is a fundamental step in predicting the outcomes of chemical reactions. Ensuring accurate measurements and conversions is vital for understanding the substance's behavior in a reaction.

Solution Concentration

Solution concentration measures the amount of solute dissolved in a solvent. It is usually expressed in molarity (M), which is moles of solute per liter of solution:

\[ C = \frac{n}{V} \]

After mixing \( \text{AgNO}_3 \) and \( \text{NaCl} \), we calculated the final concentrations of \( \text{Ag}^+ \) and \( \text{Cl}^- \):

• Concentration of \( \text{Ag}^+ \)
• \( [\text{Ag}^+] = \frac{5.0 \times 10^{-6} \text{ moles}}{0.150 \text{ L}} = 3.33 \times 10^{-5} \text{ M } \)


• Concentration of \( \text{Cl}^- \)
• \( [\text{Cl}^-] = \frac{1.0 \times 10^{-5} \text{ moles}}{0.150 \text{ L}} = 6.67 \times 10^{-5} \text{ M} \)

Accurate solution concentration calculations are critical to determining how substances interact when combined. These calculations help predict if a reaction will occur, given the solutes' propensity to react.

Reaction Quotient

The reaction quotient \( Q \) indicates the current state of a reacting system. It is given by the product of the concentrations of the reactants/species, each raised to the power of their stoichiometric coefficients:

\[ Q = [\text{Ag}^+][\text{Cl}^-] \]

For the problem, we calculated \( Q \) as follows:

• \( Q = (3.33 \times 10^{-5} M)(6.67 \times 10^{-5} M) = 2.22 \times 10^{-9} \)

The reaction quotient helps predict whether a reaction is at equilibrium, moving towards the products, or forming precipitates. Comparing with the solubility product constant (\( K_{sp} \)) determines if a precipitate forms in a solution.

Solubility Product Constant

The solubility product constant, \( K_{sp} \), is a measure of a sparingly soluble salt's solubility. It is the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. For \( \text{AgCl} \), \( K_{sp} \) is given by:

\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 1.7 \times 10^{-10} \]

In our problem, we compared \( Q \) and \( K_{sp} \):

• \( Q = 2.22 \times 10^{-9} \)
• \( K_{sp} (\text{AgCl}) = 1.7 \times 10^{-10} \)

Since \( Q > K_{sp} \), a precipitate of \( \text{AgCl} \) will form. \( K_{sp} \) values are crucial for understanding solubility dynamics and predicting the formation of precipitates from given ionic solutions.