Acids and Bases · ⁨الأحماض والقواعد⁩

Learning Objective 8.1.A: Calculate the values of $\mathrm{pH}$ and $\mathrm{pOH}$, based on $K_w$ and the concentration of all species present in a neutral solution of water.

• 8.1.A.1 The concentrations of hydronium ion and hydroxide ion are often reported as $\mathrm{pH}$ and $\mathrm{pOH}$, respectively.
  • Equation: $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$
  • Equation: $\mathrm{pOH} = -\log[\mathrm{OH^-}]$
  • The terms "hydrogen ion" and "hydronium ion" and the symbols $\mathrm{H^+}(aq)$ and $\mathrm{H_3O^+}(aq)$ are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and $\mathrm{H_3O^+}(aq)$ are preferred, but $\mathrm{H^+}(aq)$ is also accepted on the AP Exam.
• 8.1.A.2 Water autoionizes with an equilibrium constant $K_w$.
  • Equation: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ at 25°C
• 8.1.A.3 In pure water, $\mathrm{pH} = \mathrm{pOH}$ is called a neutral solution. At 25°C, $\mathrm{p}K_w = 14.0$ and thus $\mathrm{pH} = \mathrm{pOH} = 7.0$.
  • Equation: $\mathrm{p}K_w = 14 = \mathrm{pH} + \mathrm{pOH}$ at 25°C
• 8.1.A.4 The value of $K_w$ is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.

Learning Objective 8.1.A: Calculate the values of $\mathrm{pH}$ and $\mathrm{pOH}$, based on $K_w$ and the concentration of all species present in a neutral solution of water.

• 8.1.A.1 The concentrations of hydronium ion and hydroxide ion are often reported as $\mathrm{pH}$ and $\mathrm{pOH}$, respectively.
  • Equation: $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$
  • Equation: $\mathrm{pOH} = -\log[\mathrm{OH^-}]$
  • The terms "hydrogen ion" and "hydronium ion" and the symbols $\mathrm{H^+}(aq)$ and $\mathrm{H_3O^+}(aq)$ are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and $\mathrm{H_3O^+}(aq)$ are preferred, but $\mathrm{H^+}(aq)$ is also accepted on the AP Exam.
• 8.1.A.2 Water autoionizes with an equilibrium constant $K_w$.
  • Equation: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ at 25°C
• 8.1.A.3 In pure water, $\mathrm{pH} = \mathrm{pOH}$ is called a neutral solution. At 25°C, $\mathrm{p}K_w = 14.0$ and thus $\mathrm{pH} = \mathrm{pOH} = 7.0$.
  • Equation: $\mathrm{p}K_w = 14 = \mathrm{pH} + \mathrm{pOH}$ at 25°C
• 8.1.A.4 The value of $K_w$ is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.

Learning Objective 8.2.A: Calculate $\mathrm{pH}$ and $\mathrm{pOH}$ based on concentrations of all species in a solution of a strong acid or a strong base.

• 8.2.A.1 Molecules of a strong acid (e.g., $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HClO_4}$, $\mathrm{H_2SO_4}$, and $\mathrm{HNO_3}$) will completely ionize in aqueous solution to produce hydronium ions and the conjugate base of the acid. As such, the concentration of $\mathrm{H_3O^+}$ in a strong acid solution is equal to the initial concentration of the strong acid, and thus the $\mathrm{pH}$ of the strong acid solution is easily calculated.
• 8.2.A.2 When dissolved in solution, strong bases (e.g., group I and II hydroxides) completely dissociate to produce hydroxide ions. As such, the concentration of $\mathrm{OH^-}$ in a strong base solution is equal to the initial concentration of a group I hydroxide and double the initial concentration of a group II hydroxide, and thus the $\mathrm{pOH}$ (and $\mathrm{pH}$) of the strong base solution is easily calculated.

Learning Objective 8.3.A: Explain the relationship among $\mathrm{pH}$, $\mathrm{pOH}$, and concentrations of all species in a solution of a monoprotic weak acid or weak base.

• 8.3.A.1 Weak acids react with water to produce hydronium ions. However, only a small percentage of molecules of a weak acid will ionize in this way. Thus, the concentration of $\mathrm{H_3O^+}$ is much less than the initial concentration of the molecular acid, and the vast majority of the acid molecules remain un-ionized.
• 8.3.A.2 A solution of a weak acid involves equilibrium between an un-ionized acid and its conjugate base. The equilibrium constant for this reaction is $K_a$, often reported as $\mathrm{p}K_a$. The pH of a weak acid solution can be determined from the initial acid concentration and the $\mathrm{p}K_a$.
  • Equation: $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$
  • Equation: $\mathrm{p}K_a = -\log K_a$
• 8.3.A.3 Weak bases react with water to produce hydroxide ions in solution. However, ordinarily just a small percentage of the molecules of a weak base in solution will ionize in this way. Thus, the concentration of $\mathrm{OH^-}$ in the solution does not equal the initial concentration of the base, and the vast majority of the base molecules remain un-ionized.
• 8.3.A.4 A solution of a weak base involves equilibrium between an un-ionized base and its conjugate acid. The equilibrium constant for this reaction is $K_b$, often reported as $\mathrm{p}K_b$. The $\mathrm{pH}$ of a weak base solution can be determined from the initial base concentration and the $\mathrm{p}K_b$.
  • Equation: $K_b = \dfrac{[\mathrm{OH^-}][\mathrm{HB^+}]}{[\mathrm{B}]}$
  • Equation: $\mathrm{p}K_b = -\log K_b$
• 8.3.A.5 The percent ionization of a weak acid (or base) can be calculated from its $\mathrm{p}K_a$ ($\mathrm{p}K_b$) and the initial concentration of the acid (base). The percent ionization can also be calculated from the initial concentration of the acid (base) and the equilibrium concentration of any of the species in the equilibrium expression.
• 8.3.A.6 For any conjugate acid-base pair, the acid ionization constant and base ionization constant are related by $K_w$:
  • Equation: $K_w = K_a \times K_b$
  • Equation: $\mathrm{p}K_w = \mathrm{p}K_a + \mathrm{p}K_b$

Learning Objective 8.4.A: Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.

• 8.4.A.1 When a strong acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: $\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{H_2O}(l)$. The pH of the resulting solution may be determined from the concentration of excess reagent.
• 8.4.A.2 When a weak acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: $\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftarrows \mathrm{A^-}(aq)\ \mathrm{H_2O}(l)$. If the weak acid is in excess, then a buffer solution is formed, and the $\mathrm{pH}$ can be determined from the Henderson-Hasselbalch (H–H) equation (see 8.9.A.1). If the strong base is in excess, then the $\mathrm{pH}$ can be determined from the moles of excess hydroxide ion and the total volume of solution. If they are equimolar, then the (slightly basic) $\mathrm{pH}$ can be determined from the equilibrium represented by the equation: $\mathrm{A^-}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{HA}(aq) + \mathrm{OH^-}(aq)$.
• 8.4.A.3 When a weak base and a strong acid are mixed, they will react quantitatively in a reaction represented by the equation: $\mathrm{B}(aq) + \mathrm{H_3O^+}(aq) \rightleftarrows \mathrm{HB^+}(aq) + \mathrm{H_2O}(l)$. If the weak base is in excess, then a buffer solution is formed, and the $\mathrm{pH}$ can be determined from the H–H equation. If the strong acid is in excess, then the $\mathrm{pH}$ can be determined from the moles of excess hydronium ion and the total volume of solution. If they are equimolar, then the (slightly acidic) $\mathrm{pH}$ can be determined from the equilibrium represented by the equation: $\mathrm{HB^+}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{B}(aq) + \mathrm{H_3O^+}(aq)$.
• 8.4.A.4 When a weak acid and a weak base are mixed, they will react to an equilibrium state whose reaction may be represented by the equation: $\mathrm{HA}(aq) + \mathrm{B}(aq) \rightleftarrows \mathrm{A^-}(aq) + \mathrm{HB^+}(aq)$.

Learning Objective 8.5.A: Explain results from the titration of a mono- or polyprotic acid or base solution, in relation to the properties of the solution and its components.

• 8.5.A.1 An acid-base reaction can be carried out under controlled conditions in a titration. A titration curve, plotting $\mathrm{pH}$ against the volume of titrant added, is useful for summarizing results from a titration.
• 8.5.A.2 At the equivalence point for titrations of monoprotic acids or bases, the number of moles of titrant added is equal to the number of moles of analyte originally present. This relationship can be used to obtain the concentration of the analyte. This is the case for titrations of strong acids/bases and weak acids/bases.
• 8.5.A.3 For titrations of weak acids/bases, it is useful to consider the point halfway to the equivalence point, that is, the half-equivalence point. At this point, there are equal concentrations of each species in the conjugate acid-base pair, for example, for a weak acid $[\mathrm{HA}] = [\mathrm{A^-}]$. Because $\mathrm{pH} = \mathrm{p}K_a$ when the conjugate acid and base have equal concentrations, the $\mathrm{p}K_a$ can be determined from the $\mathrm{pH}$ at the half-equivalence point in a titration.
• 8.5.A.4 At the equivalence point, pH is determined by the major species in solution. Strong acid and strong base titrations result in neutral pH at the equivalence point. However, in titrations of weak acids (weak bases), the conjugate base of the weak acid (conjugate acid of the weak base) is present at the equivalence point and can undergo proton-transfer reactions with the surrounding water, producing basic (acidic) solutions.
• 8.5.A.5 For polyprotic acids, titration curves can be used to determine the number of acidic protons. In doing so, the major species present at any point along the curve can be identified, along with the $\mathrm{p}K_a$ associated with each proton in a weak polyprotic acid.
  • Exclusion statement: Computation of the concentration of each species present in the titration curve for polyprotic acids will not be assessed on the AP Exam. Such computations for titration of monoprotic acids are within the scope of the course (see 8.4.A.2 and 8.4.A.3), as is qualitative reasoning regarding what species are present in large versus small concentrations at any point in a titration of a polyprotic acid.

Learning Objective 8.6.A: Explain the relationship between the strength of an acid or base and the structure of the molecule or ion.

• 8.6.A.1 The protons on a molecule that will participate in acid-base reactions, and the relative strength of these protons, can be inferred from the molecular structure.
  • i. Strong acids (such as $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HClO_4}$, $\mathrm{H_2SO_4}$, and $\mathrm{HNO_3}$) have very weak conjugate bases that are stabilized by electronegativity, inductive effects, resonance, or some combination thereof.
  • ii. Carboxylic acids are one common class of weak acid.
  • iii. Strong bases (such as group I and II hydroxides) have very weak conjugate acids.
  • iv. Common weak bases include nitrogenous bases such as ammonia as well as carboxylate ions.
  • v. Electronegative elements tend to stabilize the conjugate base relative to the conjugate acid, and so increase acid strength.

Learning Objective 8.8.A: Explain the relationship between the ability of a buffer to stabilize $\mathrm{pH}$ and the reactions that occur when an acid or a base is added to a buffered solution.

• 8.8.A.1 A buffer solution contains a large concentration of both members in a conjugate acid-base pair. The conjugate acid reacts with added base and the conjugate base reacts with added acid. These reactions are responsible for the ability of a buffer to stabilize $\mathrm{pH}$.

Learning Objective 8.9.A: Identify the $\mathrm{pH}$ of a buffer solution based on the identity and concentrations of the conjugate acid-base pair used to create the buffer.

• 8.9.A.1 The $\mathrm{pH}$ of the buffer is related to the $\mathrm{p}K_a$ of the acid and the concentration ratio of the conjugate acid-base pair. This relation is a consequence of the equilibrium expression associated with the dissociation of a weak acid, and is described by the Henderson-Hasselbalch equation. Adding small amounts of acid or base to a buffered solution does not significantly change the ratio of $[\mathrm{A^-}]/[\mathrm{HA}]$ and thus does not significantly change the solution $\mathrm{pH}$. The change in $\mathrm{pH}$ on addition of acid or base to a buffered solution is therefore much less than it would have been in the absence of the buffer.
  • Equation: $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$
  • Exclusion statement: Computation of the change in pH resulting from the addition of an acid or a base to a buffer will not be assessed on the AP Exam.
  • Exclusion statement: Derivation of the Henderson-Hasselbalch equation will not be assessed on the AP Exam.

Learning Objective 8.10.A: Explain the relationship between the buffer capacity of a solution and the relative concentrations of the conjugate acid and conjugate base components of the solution.

• 8.10.A.1 Increasing the concentration of the buffer components (while keeping the ratio of these concentrations constant) keeps the $\mathrm{pH}$ of the buffer the same but increases the capacity of the buffer to neutralize added acid or base.
• 8.10.A.2 When a buffer has more conjugate acid than base, it has a greater buffer capacity for addition of added base than acid. When a buffer has more conjugate base than acid, it has a greater buffer capacity for addition of added acid than base.

Learning Objective 8.11.A: Identify the qualitative effect of changes in pH on the solubility of a salt.

• 8.11.A.1 The solubility of a salt is pH sensitive when one of the constituent ions is a weak acid, a weak base, or the hydroxide ion. These effects can be understood qualitatively using Le Châtelier's principle.
  • Exclusion statement: Computations of solubility as a function of pH will not be assessed on the AP Exam.