Chemistry Lecture Notes
Titrations
Key Concept: Titrations are precise analytical chemistry techniques performed to reach an equivalent point, where the number of moles of acid precisely equals the number of moles of base according to their stoichiometry. The endpoint is the visually observed point, typically indicated by a distinct color change of an indicator.
Endpoint of Titration: The point in a titration where the indicator changes color, signifying that equimolar amounts of reactants have theoretically reacted. This is an approximation of the equivalent point.
Equation for Titration Problems: For a 1:1 mole ratio reaction between acid and base, the fundamental relationship used is n\text{acid} = n\text{base} , where n represents the number of moles.
Example: To determine the unknown concentration of an acid, consider a scenario where 12.54\ \text{mL} of a standard base solution (e.g., 0.100\ \text{M NaCl}) reacts in a 1:1 stoichiometric ratio with an unknown acid (e.g., \text{HCl} ).
Given Data: For instance, a precise volume of 0.001\ \text{L} of a base with a known molarity of 0.100\ \text{M}.
Required Data: The primary objective is to accurately determine both the volume of the acid consumed at the endpoint and, subsequently, its unknown concentration.
Steps to Solving Titration Problems
Identify and Strategize: Clearly identify all given data, such as volumes and concentrations of known species, and precisely determine what unknown quantity (e.g., concentration of the analyte) needs to be calculated.
Write the Balanced Chemical Equation: This step is crucial for establishing the correct stoichiometric relationship (mole ratio) between the acid and the base. For example, if dealing with sulfuric acid ( \text{H}2\text{SO}4 ) and sodium hydroxide ( \text{NaOH} ), the balanced equation would be: \text{H}2\text{SO}4\ (aq) + 2\text{NaOH}\ (aq) \rightarrow \text{Na}2\text{SO}4\ (aq) + 2\text{H}_2\text{O}\ (l) .
Calculate the Number of Moles of the Known Reactant: Utilize the known molarity ( \text{M} ) and measured volume ( \text{L} ) of the standard solution to calculate the moles of that reactant using the formula:
\text{Moles of Known Reactant} = \text{Molarity (M)} \times \text{Volume (L)}
Subsequently, use the stoichiometric ratio from the balanced equation to calculate the moles of the unknown reactant (e.g., moles of the acid).
Find the Concentration of the Unknown Reactant: Once the moles of the unknown reactant and its titrated volume are known, its concentration ( C ) can be determined using the formula:
C = \frac{\text{Moles of Unknown Reactant}}{\text{Volume of Unknown Reactant (L)}}
Example Case: Lead Contaminated Water Titration
Context: This example involves assessing lead contamination in water samples, a critical environmental analysis. Lead (II) ions ( \text{Pb}^{2+} ) are known to be highly toxic.
Techniques Used: The analytical technique employed is complexometric titration, specifically using ethylenediaminetetraacetic acid (EDTA). EDTA is a hexadentate ligand capable of forming stable, 1:1 stoichiometric complexes with many metal ions, including \text{Pb}^{2+} .
Given: The endpoint of the titration was precisely reached with 77\ \mu\text{L} of a 5.0 \times 10^{-4}\ \text{M EDTA solution}. A conversion is necessary for microliters to liters: 1\ \mu\text{L} = 10^{-6}\ \text{L} .
Objective: The ultimate goal is to determine the concentration of lead in the water sample, expressed in micrograms per liter ( \mu\text{g/L} ).
Steps to Solve (Illustrative Calculation):
Convert EDTA volume to liters: 77\ \mu\text{L} = 77 \times 10^{-6}\ \text{L} .
Calculate moles of EDTA used: \text{Moles of EDTA} = (5.0 \times 10^{-4}\ \text{mol/L}) \times (77 \times 10^{-6}\ \text{L}) = 3.85 \times 10^{-8}\ \text{mol} .
Since EDTA reacts 1:1 with \text{Pb}^{2+} , moles of \text{Pb}^{2+} = moles of EDTA = 3.85 \times 10^{-8}\ \text{mol} .
Calculate mass of lead (molar mass of Pb = 207.2\ \text{g/mol}): \text{Mass of Pb} = (3.85 \times 10^{-8}\ \text{mol}) \times (207.2\ \text{g/mol}) = 7.98 \times 10^{-6}\ \text{g} .
Convert mass to micrograms: 7.98 \times 10^{-6}\ \text{g} = 7.98\ \mu\text{g} .
If the water sample volume was, for example, 100\ \text{mL} (or 0.1\ \text{L}), then concentration in \mu\text{g/L} would be: \frac{7.98\ \mu\text{g}}{0.1\ \text{L}} = 79.8\ \mu\text{g/L} .
Gravimetric Analysis
Methodology: Gravimetric analysis is a highly accurate quantitative analytical method for determining the amount of a substance by precisely measuring its mass. This is typically achieved by precipitating the analyte from a solution in a highly pure, insoluble form, filtering, washing, drying, and then weighing the precipitate.
Example: Silver extraction from ore. This industrial process exemplifies gravimetric analysis:
Starting with 1.397\ \text{g} of an ore sample, which is then dissolved in nitric acid ( \text{HNO}3 ) to convert any silver present into soluble silver nitrate ( \text{AgNO}3 ).
Following this, a solution of sodium chloride ( \text{NaCl} ) is added. This addition causes the precipitation of silver chloride ( \text{AgCl} ) due to its extremely low solubility, according to the reaction: \text{Ag}^{+}\ (aq) + \text{Cl}^{-}\ (aq) \rightarrow \text{AgCl}\ (s) .
The \text{AgCl} precipitate is then isolated, washed, dried, and its final mass is measured, for example, at 0.4711\ \text{g}. The high precision of this method is a key advantage.
Calculating the percentage of silver in the ore (using the given data):
Calculate moles of silver chloride ( \text{AgCl} ).
Molar mass of \text{AgCl} = 107.9\ \text{g/mol (Ag)} + 35.45\ \text{g/mol (Cl)} = 143.35\ \text{g/mol} .
n(\text{AgCl}) = \frac{0.4711\ \text{g}}{143.35\ \text{g/mol}} = 0.003286\ \text{mol} .From the stoichiometry of the precipitation reaction ( \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} ), one mole of \text{AgCl} corresponds to one mole of Ag. Therefore, moles of Ag = 0.003286\ \text{mol} .
Compute the mass of silver (Ag) present in the ore.
Mass of Ag = n(\text{Ag}) \times \text{Molar mass of Ag} = 0.003286\ \text{mol} \times 107.9\ \text{g/mol} = 0.3546\ \text{g} .Calculate the percentage of silver in the original ore sample.
Percentage of Ag = (\frac{\text{Mass of Ag}}{\text{Mass of Ore}}) \times 100\% = (\frac{0.3546\ \text{g}}{1.397\ \text{g}}) \times 100\% = 25.38\% .
Redox Reactions
Definition: Redox reactions (reduction-oxidation reactions) are chemical processes involving the transfer of electrons between two species. During such a reaction, one species undergoes oxidation (loses electrons and its oxidation state increases), while the other undergoes reduction (gains electrons and its oxidation state decreases).
Mnemonic to remember: OIL RIG (Oxidation Is Loss [of electrons], Reduction Is Gain [of electrons]).
Oxidation state basics:
For elements within a compound, the oxidation state represents the hypothetical charge an atom would have if all bonds were purely ionic and shared electrons were assigned to the more electronegative atom.
Example: Carbon can exist in a wide range of oxidation states, from -4 (as in methane, \text{CH}4 ) to +4 (as in carbon dioxide, \text{CO}2 ).
Rules to determine oxidation states:
Free elements (e.g., \text{O}2 , \text{Na} , \text{Cl}2 ) always have an oxidation state of zero.
For compounds, the sum of the oxidation states of all atoms must equal the overall charge of the compound. For neutral molecules, the sum is zero.
Common states are assigned based on known valences of specific elements:
Alkali metals (Group 1) are always +1 .
Alkaline earth metals (Group 2) are always +2 .
Hydrogen (H) is typically +1 (except in metal hydrides, where it's -1 , e.g., \text{NaH} ).
Oxygen (O) is usually -2 (except in peroxides, e.g., \text{H}2\text{O}2 , where it's -1 ).
Halogens (Group 17) are typically -1 when in compounds with less electronegative elements (e.g., \text{NaCl} ), but can have positive oxidation states with oxygen.
Determining Oxidation States Examples
Ammonia ( \text{NH}_3 ) has nitrogen with an oxidation state of -3 .
Calculation: Let the oxidation state of N be x . H is +1 . For a neutral molecule, x + 3(+1) = 0 \Rightarrow x = -3 .
For chlorine in chlorine dioxide ( \text{ClO}_2 ):
Calculation: Assuming \text{ClO}_2 is a neutral molecule and oxygen is -2 .
Let Cl be x . x + 2(-2) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4 .
This demonstrates that chlorine can exhibit various positive oxidation states depending on the compound.
Balancing Redox Reactions
Full Steps for balancing redox reactions, typically using the half-reaction method:
Separate into Half-Reactions: Identify the species that are being oxidized and reduced, and write separate unbalanced half-reactions for each. For example, for \text{Mg} + \text{O}2 \rightarrow \text{MgO} , separate into \text{Mg} \rightarrow \text{Mg}^{2+} and \text{O}2 \rightarrow \text{O}^{2-} .
Balance Atoms (Other than O and H): Ensure that atoms other than oxygen and hydrogen are balanced in each half-reaction.
Balance Oxygen Atoms: Add water molecules ( \text{H}_2\text{O} ) to the side of the equation that is deficient in oxygen.
Balance Hydrogen Atoms: For acidic solutions, add hydrogen ions ( \text{H}^+ ) to the side deficient in hydrogen. If the solution is basic, add \text{H}_2\text{O} to the side deficient in hydrogen and an equal number of \text{OH}^- to the opposite side.
Balance Charge: Add electrons ( \text{e}^- ) to the more positive side of each half-reaction to balance the total charge. The total charge on both sides of each half-reaction should be equal.
Equalize Electrons: Multiply one or both half-reactions by appropriate integer coefficients so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
Combine Half-Reactions: Add the two balanced half-reactions together. The electrons should cancel out on both sides.
Simplify: Cancel out any identical species that appear on both sides of the overall equation (e.g., \text{H}_2\text{O} , \text{H}^+ ).
Example Redox Reaction: Magnesium with Oxygen
Reaction: 2\text{Mg}\ (s) + \text{O}_2\ (g) \rightarrow 2\text{MgO}\ (s)
Oxidation states shift:
Magnesium ( \text{Mg} ) changes from an oxidation state of 0 (in its elemental form) to +2 (in \text{MgO} ). This is oxidation (loss of 2 electrons per Mg atom).
Oxidation Half-Reaction: \text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^-
Oxygen ( \text{O}2 ) changes from an oxidation state of 0 (in its elemental form) to -2 (in \text{MgO} ). This is reduction (gain of 2 electrons per O atom, or 4 electrons for \text{O}2 ).
Reduction Half-Reaction: \text{O}_2 + 4\text{e}^- \rightarrow 2\text{O}^{2-}
To balance electrons, multiply the Mg half-reaction by 2: 2\text{Mg} \rightarrow 2\text{Mg}^{2+} + 4\text{e}^- .
Combine: 2\text{Mg} + \text{O}2 + 4\text{e}^- \rightarrow 2\text{Mg}^{2+} + 2\text{O}^{2-} + 4\text{e}^- . The electrons cancel, and \text{Mg}^{2+} and \text{O}^{2-} combine to form \text{MgO} , yielding the overall balanced equation: 2\text{Mg} + \text{O}2 \rightarrow 2\text{MgO} .
In this reaction, Mg acts as the reducing agent (it gets oxidized, causing O2 to be reduced), and O2 acts as the oxidizing agent (it gets reduced, causing Mg to be oxidized).
Final Notes
The current unit will conclude with an in-depth review of electrolytes, differentiating between strong and weak electrolytes, and their conduc.tivity in solution. This will transition into the properties of gases, which comprises Unit 80, covering topics like the Ideal Gas Law ( PV=nRT ), Dalton's Law of Partial Pressures, Graham's Law of Effusion, and the Kinetic Molecular Theory.
The importance of consistent practice and achieving clarity in both practical laboratory assessments and complex theoretical calculations in chemistry is highly emphasized for successfully mastering these upcoming topics. This includes developing strong problem-solving skills and critical thinking.