Reactions in Aqueous Solution: Displacement Reactions, Activity Series, and Solution Concentrations

Chapter 4: Reactions in Aqueous Solution (Part 3)

Displacement Reactions

Displacement reactions are a type of redox reaction in which an ion in solution oxidizes an element, causing the element to be displaced (replaced) by another species. These reactions are fundamental in understanding how metals react in aqueous environments.

• Definition: A displacement reaction occurs when an element reacts with a compound, displacing another element from it.

• Example: When magnesium metal is placed in hydrochloric acid, magnesium displaces hydrogen ions, producing hydrogen gas and magnesium chloride.

General equation:

• $A + BC \rightarrow AC + B$

Redox aspect: The element that is oxidized loses electrons, while the ion that is reduced gains electrons.

• Example: $\ce{Mg(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H2(g)}$

• Here, $\ce{Mg}$ is oxidized (loses electrons), and $\ce{H^+}$ is reduced (gains electrons).

Activity Series and Hydrogen

The activity series is a list of elements organized by their ability to be oxidized (lose electrons). It helps predict whether a displacement reaction will occur.

• Elements above hydrogen in the activity series will react with acids to produce hydrogen gas.

• Elements below hydrogen will not react with acids to produce hydrogen gas.

• A reactive metal (higher in the series) can displace a less reactive metal (lower in the series) from its compounds.

Table: Activity Series of Metals in Aqueous Solution

Metal	Oxidation Reaction
Li	$\ce{Li -> Li^+ + e^-}$
K	$\ce{K -> K^+ + e^-}$
Ca	$\ce{Ca -> Ca^{2+} + 2e^-}$
Na	$\ce{Na -> Na^+ + e^-}$
Mg	$\ce{Mg -> Mg^{2+} + 2e^-}$
Al	$\ce{Al -> Al^{3+} + 3e^-}$
Zn	$\ce{Zn -> Zn^{2+} + 2e^-}$
Fe	$\ce{Fe -> Fe^{2+} + 2e^-}$
Pb	$\ce{Pb -> Pb^{2+} + 2e^-}$
H+	$\ce{2H^+ + 2e^- -> H2}$
Cu	$\ce{Cu -> Cu^{2+} + 2e^-}$
Ag	$\ce{Ag -> Ag^+ + e^-}$
Au	$\ce{Au -> Au^{3+} + 3e^-}$

Additional info: The table above is a partial activity series, showing that metals higher up are more easily oxidized and thus more reactive in displacement reactions.

Acid Displacement Reactions

Acid displacement reactions involve a metal reacting with an acid, resulting in the release of hydrogen gas and the formation of a salt.

• Metals higher on the activity series are more reactive and will exist as ions in solution.

• Metals lower on the activity series will remain as elements and not react with acids.

• Example: $\ce{Zn(s) + 2HCl(aq) \rightarrow ZnCl2(aq) + H2(g)}$

Concentration of Solutions

The concentration of a solution is a measure of the amount of solute dissolved in a given quantity of solvent or solution. The most common unit is molarity (M).

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$

• Example: A solution contains 1.46 moles of $\ce{C6H12O6}$ in 1.00 L of solution. $M = \frac{1.46\ \text{mol}}{1.00\ \text{L}} = 1.46\ \text{M}$

Concentration of Ions in Solution

When strong electrolytes dissolve in water, they dissociate completely into ions. The concentration of each ion can be determined from the stoichiometry of the dissociation.

• Example 1: $\ce{KCl(s) \rightarrow K^+(aq) + Cl^-(aq)}$ If 1.0 M $\ce{KCl}$ is dissolved, $[\ce{K^+}] = 1.0$ M and $[\ce{Cl^-}] = 1.0$ M.

• Example 2: $\ce{Ba(NO3)2(s) \rightarrow Ba^{2+}(aq) + 2NO3^-(aq)}$ If 1.0 M $\ce{Ba(NO3)2}$ is dissolved, $[\ce{Ba^{2+}}] = 1.0$ M and $[\ce{NO3^-}] = 2.0$ M.

Preparing a Solution of Known Molarity

To prepare a solution of a specific molarity, a known mass (or number of moles) of solute is dissolved in a volumetric flask and diluted to a known volume with solvent.

• Steps:

  1. Calculate the required moles of solute: $n = M \times V$

  2. Convert moles to mass: $\text{mass} = n \times \text{molar mass}$

  3. Weigh out the solute and dissolve in a small amount of solvent.

  4. Transfer to a volumetric flask and dilute to the desired final volume.

• Example: What mass of $\ce{KI}$ is required to make 500 mL of 2.80 M $\ce{KI}$ solution?

  • Convert volume to liters: $500\ \text{mL} = 0.500\ \text{L}$

  • Calculate moles: $n = 2.80\ \text{mol/L} \times 0.500\ \text{L} = 1.40\ \text{mol}$

  • Molar mass of $\ce{KI}$ = 166 g/mol

  • Mass = $1.40\ \text{mol} \times 166\ \text{g/mol} = 232.4\ \text{g}$

Dilution of Solutions

Dilution is the process of preparing a less concentrated solution from a more concentrated one by adding solvent. The number of moles of solute remains constant before and after dilution.

• Key equation: $M_i V_i = M_f V_f$

• Example: To prepare 60.0 mL of 0.200 M $\ce{HNO3}$ from a 4.00 M stock solution:

  • $M_i = 4.00$ M, $V_f = 60.0$ mL, $M_f = 0.200$ M

  • $V_i = \frac{M_f V_f}{M_i} = \frac{0.200 \times 60.0}{4.00} = 3.00$ mL

  • Mix 3.00 mL of stock solution with 57.0 mL of water to make 60.0 mL of 0.200 M solution.

Acid-Base Titration

Titration is a technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration (standard solution). The reaction is complete at the equivalence point, where stoichiometrically equivalent amounts of acid and base have reacted.

• Indicator: A substance that changes color at (or near) the equivalence point, signaling the end of the titration.

• General reaction: $\ce{Acid + Base \rightarrow Salt + Water}$

• Example: $\ce{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H2O(l)}$

• Stoichiometry: Use the balanced equation to relate moles of acid and base.

• Calculation: $M_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}$ (for 1:1 ratio)

• Example Problem 1: 42.00 mL of 0.150 M NaOH is required to neutralize 50.00 mL of HCl. What is the molarity of the acid?

  • $M_{\text{acid}} \times 50.00 = 0.150 \times 42.00$

  • $M_{\text{acid}} = \frac{0.150 \times 42.00}{50.00} = 0.126$ M

• Example Problem 2: 42.00 mL of 0.150 M NaOH is required to neutralize 50.00 mL of $\ce{H2SO4}$. What is the molarity of the acid?

  • Balanced equation: $\ce{H2SO4 + 2NaOH \rightarrow 2H2O + Na2SO4}$

  • Stoichiometry: 1 mol $\ce{H2SO4}$ reacts with 2 mol $\ce{NaOH}$

  • $M_{\text{acid}} \times 50.00 \times 2 = 0.150 \times 42.00$

  • $M_{\text{acid}} = \frac{0.150 \times 42.00}{50.00 \times 2} = 0.0630$ M

Additional info: In titration calculations, always use the stoichiometric coefficients from the balanced equation to relate moles of acid and base.