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General Chemistry Study Guide

Chapter 4. Reactions in Aqueous Solutions


Yu Wang

1. General Properties of Aqueous Solutions

OpenStax 3.3 Molarity. Brown 4.1 General Properties of Aqueous Solutions

A solution is a homogenous mixture of 2 or more substances.

The solute is (are) the substance(s) present in the smaller amount(s).

The solvent is the substance present in the larger amount.

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. Electrolytes are generally ionic coumponds.

A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.

Requirements

  1. Understand and remember the concepts.

2. Precipitation Reactions

OpenStax 4.1 Writing and Balancing Chemical Equations; 4.2 Classifying Chemical Reactions. Brown 4.2 Precipitation Reactions.

Precipitate – insoluble solid that separates from solution

Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.

Check this website to see the solubility of common ionic coumponds.

Memorize the following:

Writing Net Ionic Equations

  1. Write the balanced molecular equation.
  2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.
  3. Cancel the spectator ions on both sides of the ionic equation.
  4. Check that charges and number of atoms are balanced in the net ionic equation.
Example: Give the ionic form and the net ionic form of the following reaction. $$\ce{AgNO3 (aq) + NaCl (aq) -> AgCl (s) + NaNO3 (aq)}$$ Answer:
In ionic form: $$\ce{Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) -> AgCl (s) + Na+ (aq) + NO3- (aq)}$$ Net ionic form: $$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)}$$
Example: Give the ionic form and the net ionic form of the following reaction. $$\ce{Ba(OH)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaOH (aq)}$$ Answer:
In ionic form: $$\ce{Ba^{2+} (aq) + 2OH- (aq) + 2Na+ (aq) + SO4^{2-} (aq) -> BaSO4 (s) + 2Na+ (aq) + 2OH- (aq)}$$ Net ionic form: $$\ce{Ba^{2+} (aq) + SO4^{2-} (aq) -> BaSO4 (s)}$$

Requirements

  1. Understand and remember the concepts.
  2. Get familiar with some insoluble salts.

3. Acid-Base Reactions

OpenStax 4.2 Classifying Chemical Reactions. Brown 4.3 Acids, Bases, and Neutralization Reactions.

Properties of acids

Properties of bases

Arrhenius acid is a substance that produces $\ce{H+}$ ($\ce{H3O+}$) in water.
$$\ce{HCl + H2O <=> H3O+ + Cl-}$$

Arrhenius base is a substance that produces $\ce{OH-}$ in water.

$$\ce{NaOH + H2O <=> OH- + Na+}$$

A Brønsted acid is a proton donor. A Brønsted acid must contain at least one ionizable proton.
A Brønsted base is a proton acceptor.

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

In the forward reaction $\ce{H2O}$ is the acid and $\ce{NH3}$ is the base; in the reverse reaction $\ce{NH4+}$ is the acid and $\ce{OH-}$ is the base.

Common strong and weak acids. Strong acids ionize completely in water; weak acids ionize partially in water.

List of common strong acids.

List of common weak acids

The last 4 examples in the list are organic acids with $\ce{-COOH}$. These organic acids are all week acids.

A nutralization reactions when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water.

Strong acid and strong base

Example: Give the net ionic form of the following reaction. $$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O}$$ Answer:
$$\ce{H+ + Cl- + Na+ + OH- -> Na+ + Cl- + H2O}$$ $$\ce{H+ + OH- -> H2O}$$

Weak acid and strong base

Example: Give the net ionic form of the following reaction. $$\ce{HCN(aq) + NaOH(aq) -> NaCN(aq) + H2O}$$ Answer:
$$\ce{HCN + Na+ + OH- -> Na+ + CN- + H2O}$$ $$\ce{HCN + OH- -> CN- + H2O}$$

Requirements

  1. Understand and remember the concepts.
  2. Remember the common strong and weak acids.
  3. Understand the difference of strong and weak acids in nutralization reactions.

4. Oxidation-Reduction Reactions

OpenStax 4.2 Classifying Chemical Reactions. Brown 4.4 Oxidation-Reduction Reactions.

Redox (short for reduction–oxidation reaction) is a chemical reaction in which the oxidation states of atoms are changed. Any such reaction involves both a reduction process and a complementary oxidation process.

Example:

In this example, $\ce{Na}$ loses one electron becoming $\ce{Na+}$. We call this process the oxidation of $\ce{Na}$. Simultaneously, $\ce{F}$ gains one electron becoming $\ce{F-}$. This process is called the reduction of $\ce{F}$.

Reducing agent is an element or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction. Since the reducing agent is losing electrons, it is said to have been oxidized.
Oxidizing agent is a substance that has the ability to oxidize other substances (cause them to lose electrons). In a redox reaction, the oxidizing agent itself is reduced.

Oxidation Number

The oxidation state, often called the oxidation number, is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component.

  1. The oxidation state of a free element (uncombined element) is zero.
  2. For a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion.
  3. Hydrogen has an oxidation state of $+1$ and oxygen has an oxidation state of $−2$ when they are present in most compounds. Exceptions to this are that hydrogen has an oxidation state of $−1$ in hydrides of active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of $−1$ in peroxides, e.g. $\ce{H2O2}$.
  4. Group IA metals have oxidation states of $+1$; IIA metals have $+2$. Fluorine always has an oxidation state of $-1$.
  5. The algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.
  6. Oxidation numbers do not have to be integers. The oxidation number of oxygen in the superoxide ion, $\ce{O2^-}$, is $-1/2$.

The oxidation numbers of elements in their compounds (upper right corner of each element):

Types of redox reactions

Combination Reaction

Combination is a reaction where two or more elements or compounds (reactants) combine to form a single compound (product).

Example

$$\ce{2Al + 3Br2 -> 2AlBr3}$$

Decomposition Reaction

Decomposition means the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.

Example

$$\ce{2KClO3 -> 2KCl + 3O2}$$

Combustion Reaction

Combustion reactions always involve molecular oxygen $\ce{O2}$. Anytime anything burns (in the usual sense), it is a combustion reaction.

Example

$$\ce{2H2 + O2 -> 2H2O}$$

Displacement Reaction

Hydrogen Displacement Reaction

Metals react with water or acid to produce hydrogen.

Example

$$\ce{Zn + 2HCl -> ZnCl2 + H2}$$

The Activity series for metals

Metal Activity
Li, K, Ba, Ca, Na React with cold water to produce $\ce{H2}$.
Mg, Al, Zn, Cr, Fe, Cd React with steam to produce $\ce{H2}$.
Co, Ni, Sn, Pb React with acids to produce $\ce{H2}$.
Cu, Ag, Hg, Pt, Au Do not react with water or acids to produce $\ce{H2}$.

Metal Displacement Reaction

More active metals displace less active metals.

Example

$$\ce{TiCl4 + 2Mg -> Ti + 2MgCl2}$$

Halogen Displacement Reaction

More active halogen produces less active halogen

$$\ce{F2 > Cl2 > Br2 > I2}$$

Example

$$\ce{Cl2 + 2KBr -> 2KCl + Br2}$$

Disproportionation Reaction

The same element is simultaneously oxidized and reduced.

Example

$$\ce{Cl2 + 2OH- -> ClO- + Cl- + H2O}$$
Example: Classify the following redox reactions and indicate changes in the oxidation numbers of the elements. \begin{align*} & \text{(a)}\,\ce{2N2O(g) -> 2N2(g) + O2(g)} \\ & \text{(b)}\,\ce{6Li(s) + N2(g) -> 2Li3N(s)} \\ & \text{(c)}\,\ce{Ni(s) + Pb(NO3)2(aq) -> Pb(s) + Ni(NO3)2(aq)} \end{align*}
Answer:
(a) Decomposition reaction. The products, $\ce{N2}$ and $\ce{O2}$, are free elements. The oxdation numbers must be 0. In the reactant, O has an oxidation number of $-2$ according to the rule, N must be $+1$. (b) Combination reaction. The reactants, $\ce{Li}$ and $\ce{N2}$, are free elements. The oxdation numbers must be 0. In the product, Li has an oxidation number of $+1$ according to the rule, N must be $-3$. (c) Metal displacement reaction. $\ce{NO3-}$ anion does not change in this reaction. According to the rule, O has an oxidation number of $-2$, thus N must be $+5$. In the reactants, Ni as free element has an oxidation number 0, Pb must be $+2$ to make $\ce{Pb(NO3)2}$ neutral. In the products, Pb as free element has an oxidation number 0, Ni must be $+2$ to make $\ce{Pb(NO3)2}$ neutral.

Requirements

  1. Understand and remember the concepts.
  2. Know how to determine oxidation numbers.
  3. Remember the relative activities of metals. Alkali metals plus Ca, Ba are very active and can react with cool water. Mg, Al, Zn, Fe can react with steam. Ni and Pb only react with acid. Cu, Ag, Hg, Au, Pt do not react with acid to produce hydrogen.
  4. Remember at lease one example of each type of redox reactions.

5. Solution Stoichiometry

OpenStax 3.3 Molarity; 4.5 Quantitative Chemical Analysis. Brown 4.5 Concentrations of Solutions; 4.6 Solution Stoichiometry and Chemical Analysis.

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

$$M = \text{molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}$$

The unit is mol/L. You can also use M as the unit, which is the same as mol/L.

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

Moles of solute does not change

Thus, $$M_iV_i = M_fV_f$$

Example: To make a solution of glucose ($\ce{C6H12O6}$) with the concentration of 0.283 mol/L and volume of 250 mL. How many grams of glucose is needed? Answer:
$$\text{molar mass of glucose} = 6\times 12.01\,\text{g/mol} + 12\times 1.01\,\text{g/mol} + 6\times 16.00\,\text{g/mol} = 180.18\,\text{g/mol}$$ $$\text{mass of glucose} = 250\,\text{mL}\times\frac{1\,\text{L}}{1000\,\text{mL}}\times\frac{0.283\,\text{mol}}{1\,\text{L}}\times\frac{180.18\,\text{g}}{1\,\text{mol}}=12.8\,\text{g}$$
Example: How you would prepare 50 mL of a 1.00 mol/L HCl solution, starting with an 3.92 mol/L stock solution of HCl? Answer:
Take a certain volume ($V_i$) of 3.92 mol/L HCl, and add water till the overall final volume ($V_f$) is 50 mL. To caculate $V_i$, we use the equation $$M_i\times V_i = M_f \times V_f$$ Thus, $$V_i = \frac{M_f \times V_f}{M_i} = \frac{1.00\,\text{mol/L}\times\frac{1\,\text{L}}{1000\,\text{mL}}\times 50\,\text{mL}}{3.92\,\text{mol/L}}=0.0128\,\text{L}=12.8\,\text{mL}$$

Gravimetric Analysis

  1. Dissolve unknown substance in water
  2. React unknown with known substance to form a precipitate
  3. Filter and dry precipitate
  4. Weigh precipitate
  5. Use chemical formula and mass of precipitate to determine amount of unknown ion

Titrations

In a titration, a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Titrations can be used in the analysis of acid-base reactions and redox reactions.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Requirements

  1. Understand and remember the concepts. Remember the defination of molarity.
  2. Learn how to do calculations in gravimetric analysis and in titrations.

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