All Study Guides Inorganic Chemistry I Unit 6
🧶 Inorganic Chemistry I Unit 6 – Acid-Base Concepts in Inorganic ChemAcid-base concepts form the foundation of inorganic chemistry, explaining how substances interact and react. These principles cover everything from simple proton transfers to complex buffer systems, providing a framework for understanding chemical behavior in various environments.
From Arrhenius to Lewis theories, acid-base concepts have evolved to encompass a wide range of chemical interactions. Understanding pH, neutralization reactions, and buffer solutions is crucial for predicting and controlling chemical processes in both laboratory and real-world applications.
Key Concepts and Definitions
Acids produce hydrogen ions (H + H^+ H + ) in aqueous solutions
Bases produce hydroxide ions (O H − OH^- O H − ) in aqueous solutions
Amphoteric substances can act as either an acid or a base depending on the reaction conditions (water)
Conjugate acid-base pairs consist of two species that differ by a proton (H + H^+ H + ) (N H 4 + NH_4^+ N H 4 + and N H 3 NH_3 N H 3 )
The conjugate base is formed by removing a proton from the acid
The conjugate acid is formed by adding a proton to the base
Neutralization reactions involve the combination of an acid and a base to form water and a salt
Autoionization of water refers to the self-ionization of water molecules into H + H^+ H + and O H − OH^- O H − ions
The ion product constant of water (K w K_w K w ) is equal to [ H + ] [ O H − ] = 1.0 × 1 0 − 14 [H^+][OH^-] = 1.0 \times 10^{-14} [ H + ] [ O H − ] = 1.0 × 1 0 − 14 at 25°C
Acid-Base Theories
Arrhenius theory defines acids as substances that dissociate in water to produce H + H^+ H + ions and bases as substances that dissociate to produce O H − OH^- O H − ions
Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors
Focuses on the transfer of protons (H + H^+ H + ) between species
Allows for the identification of conjugate acid-base pairs
Lewis theory defines acids as electron pair acceptors and bases as electron pair donors
Broadens the definition of acids and bases beyond proton transfer
Includes species that do not contain protons (B F 3 BF_3 B F 3 , A l C l 3 AlCl_3 A lC l 3 )
Solvent system definition considers the role of the solvent in acid-base reactions
In non-aqueous solvents, the solvent can act as an acid or base (liquid ammonia, glacial acetic acid)
Usanovich theory combines the Lewis and Brønsted-Lowry theories
Defines acids as species that accept negative species or donate positive species
Defines bases as species that accept positive species or donate negative species
Types of Acids and Bases
Strong acids completely dissociate in aqueous solutions (H C l HCl H Cl , H N O 3 HNO_3 H N O 3 , H 2 S O 4 H_2SO_4 H 2 S O 4 )
Have a large acid dissociation constant (K a K_a K a )
Produce a high concentration of H + H^+ H + ions in solution
Weak acids partially dissociate in aqueous solutions (C H 3 C O O H CH_3COOH C H 3 COO H , H F HF H F , H 2 C O 3 H_2CO_3 H 2 C O 3 )
Have a small acid dissociation constant (K a K_a K a )
Establish an equilibrium between the acid and its dissociated ions
Strong bases completely dissociate in aqueous solutions (N a O H NaOH N a O H , K O H KOH K O H )
Produce a high concentration of O H − OH^- O H − ions in solution
Weak bases partially dissociate in aqueous solutions (N H 3 NH_3 N H 3 , C H 3 N H 2 CH_3NH_2 C H 3 N H 2 )
Establish an equilibrium between the base and its dissociated ions
Polyprotic acids can donate multiple protons (H 3 P O 4 H_3PO_4 H 3 P O 4 , H 2 S O 4 H_2SO_4 H 2 S O 4 )
Each successive proton dissociation has a smaller acid dissociation constant (K a 1 > K a 2 > K a 3 K_{a1} > K_{a2} > K_{a3} K a 1 > K a 2 > K a 3 )
Organic acids contain a carboxyl group (− C O O H -COOH − COO H ) and can act as weak acids (benzoic acid, citric acid)
Acid-Base Reactions
Neutralization reactions involve the combination of an acid and a base to form water and a salt
The net ionic equation for a strong acid-strong base reaction is: H + ( a q ) + O H − ( a q ) → H 2 O ( l ) H^+(aq) + OH^-(aq) \rightarrow H_2O(l) H + ( a q ) + O H − ( a q ) → H 2 O ( l )
The pH at the equivalence point of a strong acid-strong base titration is 7
Acid-base titrations are used to determine the concentration of an unknown acid or base solution
Involves the gradual addition of a titrant (acid or base of known concentration) to the analyte (unknown solution)
The equivalence point is reached when the moles of titrant equal the moles of analyte
Hydrolysis reactions involve the reaction of a salt with water to produce an acidic or basic solution
Salts of weak acids and strong bases produce basic solutions (sodium acetate)
Salts of strong acids and weak bases produce acidic solutions (ammonium chloride)
Acid-base indicators are weak acids or bases that change color depending on the pH of the solution
The color change occurs over a specific pH range, allowing for the visual determination of pH
Common indicators include phenolphthalein, methyl orange, and bromothymol blue
Lewis acid-base reactions involve the formation of a coordinate covalent bond
The Lewis acid accepts an electron pair from the Lewis base (B F 3 BF_3 B F 3 + N H 3 NH_3 N H 3 → \rightarrow → F 3 B − N H 3 F_3B-NH_3 F 3 B − N H 3 )
pH and pOH
pH is a logarithmic scale that measures the acidity of a solution
Defined as the negative logarithm of the hydrogen ion concentration: p H = − log [ H + ] pH = -\log[H^+] p H = − log [ H + ]
A pH less than 7 indicates an acidic solution, while a pH greater than 7 indicates a basic solution
pOH is a logarithmic scale that measures the basicity of a solution
Defined as the negative logarithm of the hydroxide ion concentration: p O H = − log [ O H − ] pOH = -\log[OH^-] pO H = − log [ O H − ]
A pOH less than 7 indicates a basic solution, while a pOH greater than 7 indicates an acidic solution
The relationship between pH and pOH is given by: p H + p O H = 14 pH + pOH = 14 p H + pO H = 14
The pH of a solution can be calculated from the concentration of H + H^+ H + ions
For a strong acid: p H = − log [ H + ] = − log ( C a ) pH = -\log[H^+] = -\log(C_a) p H = − log [ H + ] = − log ( C a ) , where C a C_a C a is the concentration of the acid
For a weak acid: p H = − log [ H + ] = − log ( K a C a ) pH = -\log[H^+] = -\log(\sqrt{K_aC_a}) p H = − log [ H + ] = − log ( K a C a ) , where K a K_a K a is the acid dissociation constant
The pOH of a solution can be calculated from the concentration of O H − OH^- O H − ions
For a strong base: p O H = − log [ O H − ] = − log ( C b ) pOH = -\log[OH^-] = -\log(C_b) pO H = − log [ O H − ] = − log ( C b ) , where C b C_b C b is the concentration of the base
For a weak base: p O H = − log [ O H − ] = − log ( K b C b ) pOH = -\log[OH^-] = -\log(\sqrt{K_bC_b}) pO H = − log [ O H − ] = − log ( K b C b ) , where K b K_b K b is the base dissociation constant
Buffers and Buffer Solutions
Buffers are solutions that resist changes in pH when small amounts of acid or base are added
Consist of a weak acid and its conjugate base, or a weak base and its conjugate acid
Maintain a relatively constant pH through the consumption of added H + H^+ H + or O H − OH^- O H − ions
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of the acid and its conjugate base
For an acidic buffer: p H = p K a + log [ A − ] [ H A ] pH = pK_a + \log\frac{[A^-]}{[HA]} p H = p K a + log [ H A ] [ A − ] , where p K a = − log K a pK_a = -\log K_a p K a = − log K a
For a basic buffer: p O H = p K b + log [ B ] [ B H + ] pOH = pK_b + \log\frac{[B]}{[BH^+]} pO H = p K b + log [ B H + ] [ B ] , where p K b = − log K b pK_b = -\log K_b p K b = − log K b
The buffer capacity is a measure of the ability of a buffer to resist changes in pH
Depends on the concentrations of the acid and its conjugate base, as well as the ratio of these concentrations
A higher buffer capacity indicates a greater ability to maintain a stable pH
Buffers are important in biological systems to maintain the pH of bodily fluids (blood, cytoplasm)
The bicarbonate buffer system (H 2 C O 3 H_2CO_3 H 2 C O 3 /H C O 3 − HCO_3^- H C O 3 − ) is a major buffer in blood
Phosphate buffers (H 2 P O 4 − H_2PO_4^- H 2 P O 4 − /H P O 4 2 − HPO_4^{2-} H P O 4 2 − ) are important in intracellular fluids
Buffers are also used in chemical processes to control the pH of reactions
Maintain optimal conditions for enzyme activity, product formation, or separation processes
Applications in Inorganic Chemistry
Acid-base reactions are involved in the synthesis and purification of inorganic compounds
Precipitation reactions rely on the formation of insoluble salts from the reaction of acids and bases (B a C l 2 + H 2 S O 4 → B a S O 4 ↓ + 2 H C l BaCl_2 + H_2SO_4 \rightarrow BaSO_4 \downarrow + 2HCl B a C l 2 + H 2 S O 4 → B a S O 4 ↓ + 2 H Cl )
Acid-base extraction is used to separate compounds based on their solubility in acidic or basic solutions
Acid-base properties influence the solubility and reactivity of inorganic compounds
Amphoteric oxides (A l 2 O 3 Al_2O_3 A l 2 O 3 , Z n O ZnO Z n O ) can react with both acids and bases
Basic oxides (N a 2 O Na_2O N a 2 O , C a O CaO C a O ) react with acids to form salts and water
Acid-base reactions are involved in the corrosion of metals
Acidic environments promote the oxidation of metals and the formation of soluble metal ions
Basic environments can lead to the formation of protective oxide layers on metal surfaces
Acid-base chemistry is important in environmental and industrial processes
The pH of soil and water affects the availability of nutrients and the survival of organisms
Acid rain, caused by the dissolution of atmospheric pollutants (S O 2 SO_2 S O 2 , N O x NO_x N O x ), can harm ecosystems
The pH control of industrial wastewater is necessary to meet environmental regulations and prevent equipment damage
Acid-base reactions are used in analytical chemistry techniques
Acid-base titrations are used to determine the concentration of acids or bases in solutions
pH indicators are used to visually detect the endpoint of titrations or to estimate the pH of solutions
Problem-Solving and Calculations
Calculating the pH or pOH of a solution from the concentration of H + H^+ H + or O H − OH^- O H − ions
Example: Calculate the pH of a 0.025 M H C l HCl H Cl solution
p H = − log [ H + ] = − log ( 0.025 ) = 1.60 pH = -\log[H^+] = -\log(0.025) = 1.60 p H = − log [ H + ] = − log ( 0.025 ) = 1.60
Calculating the concentration of H + H^+ H + or O H − OH^- O H − ions from the pH or pOH of a solution
Example: Calculate the [ O H − ] [OH^-] [ O H − ] in a solution with a pH of 9.5
p O H = 14 − p H = 14 − 9.5 = 4.5 pOH = 14 - pH = 14 - 9.5 = 4.5 pO H = 14 − p H = 14 − 9.5 = 4.5
[ O H − ] = 1 0 − p O H = 1 0 − 4.5 = 3.2 × 1 0 − 5 M [OH^-] = 10^{-pOH} = 10^{-4.5} = 3.2 \times 10^{-5} M [ O H − ] = 1 0 − pO H = 1 0 − 4.5 = 3.2 × 1 0 − 5 M
Calculating the pH of a weak acid or weak base solution using the acid or base dissociation constant
Example: Calculate the pH of a 0.1 M C H 3 C O O H CH_3COOH C H 3 COO H solution (K a = 1.8 × 1 0 − 5 K_a = 1.8 \times 10^{-5} K a = 1.8 × 1 0 − 5 )
p H = − log [ H + ] = − log ( K a C a ) = − log ( ( 1.8 × 1 0 − 5 ) ( 0.1 ) ) = 2.87 pH = -\log[H^+] = -\log(\sqrt{K_aC_a}) = -\log(\sqrt{(1.8 \times 10^{-5})(0.1)}) = 2.87 p H = − log [ H + ] = − log ( K a C a ) = − log ( ( 1.8 × 1 0 − 5 ) ( 0.1 ) ) = 2.87
Calculating the pH of a buffer solution using the Henderson-Hasselbalch equation
Example: Calculate the pH of a buffer containing 0.2 M C H 3 C O O H CH_3COOH C H 3 COO H and 0.1 M C H 3 C O O N a CH_3COONa C H 3 COON a (K a = 1.8 × 1 0 − 5 K_a = 1.8 \times 10^{-5} K a = 1.8 × 1 0 − 5 )
p H = p K a + log [ A − ] [ H A ] = − log ( 1.8 × 1 0 − 5 ) + log 0.1 0.2 = 4.74 + ( − 0.30 ) = 4.44 pH = pK_a + \log\frac{[A^-]}{[HA]} = -\log(1.8 \times 10^{-5}) + \log\frac{0.1}{0.2} = 4.74 + (-0.30) = 4.44 p H = p K a + log [ H A ] [ A − ] = − log ( 1.8 × 1 0 − 5 ) + log 0.2 0.1 = 4.74 + ( − 0.30 ) = 4.44
Solving acid-base titration problems to determine the concentration of an unknown solution
Example: A 25.0 mL sample of H C l HCl H Cl is titrated with 0.100 M N a O H NaOH N a O H . The equivalence point is reached after 20.0 mL of N a O H NaOH N a O H is added. Calculate the concentration of the H C l HCl H Cl solution.
At the equivalence point: n ( H C l ) = n ( N a O H ) n(HCl) = n(NaOH) n ( H Cl ) = n ( N a O H )
M ( H C l ) × V ( H C l ) = M ( N a O H ) × V ( N a O H ) M(HCl) \times V(HCl) = M(NaOH) \times V(NaOH) M ( H Cl ) × V ( H Cl ) = M ( N a O H ) × V ( N a O H )
M ( H C l ) = M ( N a O H ) × V ( N a O H ) V ( H C l ) = ( 0.100 M ) ( 20.0 m L ) 25.0 m L = 0.0800 M M(HCl) = \frac{M(NaOH) \times V(NaOH)}{V(HCl)} = \frac{(0.100 M)(20.0 mL)}{25.0 mL} = 0.0800 M M ( H Cl ) = V ( H Cl ) M ( N a O H ) × V ( N a O H ) = 25.0 m L ( 0.100 M ) ( 20.0 m L ) = 0.0800 M