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Chapter 16: Problem 42

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

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

Step by step solution

01

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 \(\text{NaCl}\:\) \[ n_{\text{NaCl}} = 1.0 \times 10^{-4} M \times 0.100 L = 1.0 \times 10^{-5} \text{ moles} \]
02

Calculate Final Volumes

Add the volumes of \(\text{AgNO}_3\) and \(\text{NaCl}\): \[ V_{\text{total}} = 50.0 \text{ mL} + 100.0 \text{ mL} = 150.0 \text{ mL} = 0.150 \text{ L} \]
03

Calculate Final Concentrations

Determine the final concentrations of \(\text{Ag}^+\) and \(\text{Cl}^-\) in the mixed solution: \[ [\text{Ag}^+] = \frac{5.0 \times 10^{-6} \text{ moles}}{0.150 \text{ L}} = 3.33 \times 10^{-5} M \]\[ [\text{Cl}^-] = \frac{1.0 \times 10^{-5} \text{ moles}}{0.150 \text{ L}} = 6.67 \times 10^{-5} M \]
04

Calculate the Reaction Quotient Q

Calculate the reaction quotient \(Q\) to determine if a precipitate will form: \[ Q = [\text{Ag}^+][\text{Cl}^-] \]\[ Q = (3.33 \times 10^{-5} M)(6.67 \times 10^{-5} M) = 2.22 \times 10^{-9} \]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Suppose the concentration of a solution is \(0.10 M \mathrm{Ba}^{2+}\) and \(0.10 \mathrm{M} \mathrm{Sr}^{2+}\). Which sulfate, \(\mathrm{BaSO}_{4}\) or \(\mathrm{SrSO}_{4}\), will precipitate first when a dilute solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is added dropwise to the solution? Show evidence for your answer. \(\left(K_{\mathrm{sp}}=1.5 \times 10^{-9}\right.\) for \(\mathrm{BaSO}_{4}\) and \(K_{\mathrm{sp}}=3.5 \times 10^{-7}\) for \(\mathrm{SrSO}_{4}\) )

Sketch the energy diagram for an endothermic reaction using a solid line. Sketch what would happen to the energy profile if a catalyst is added to the reaction using a dotted line.

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Calculate the molar solubility for these substances: (a) \(\mathrm{PbSO}_{4}, K_{\mathrm{sp}}=1.3 \times 10^{-8}\) (b) \(\mathrm{BaCrO}_{4}, K_{\mathrm{sp}}=8.5 \times 10^{-11}\)
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