Salt Hydrolysis

Fundamentals of Salt Formation and Neutralization

• Definition of a Salt: Salts are ionic compounds formed when the hydrogen ion of an acid is replaced by a metal cation or an ammonium/alkylammonium cation.

• General Reaction of Neutralization: Acids and bases react to produce a salt and water:   $\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$

• Water Formation in Neutralization: In reactions such as nitric acid reacting with sodium hydroxide, the $\text{H}^+$ ion from the acid combines with the $\text{OH}^-$ ion from the base to yield water ($\text{H}_2\text{O}$, sometimes written as $\text{HOH}$ to illustrate that one hydrogen originates from the acid and the hydroxyl group originates from the base). The remaining anion from the acid and cation from the base combine to form the salt:   $\text{HNO}_3(aq) + \text{NaOH}(aq) \rightarrow \text{HOH}(l) + \text{NaNO}_3(aq)$   In this reaction, $\text{NaNO}_3(aq)$ is the salt sodium nitrate.

• Identifying Salts: Any compound containing an anion known to originate from an acid combined with a metal or non-acidic cation is classified as a salt:

  • Acetate anion ($\text{CH}_3\text{COO}^-$) from ethanoic acid ($\text{CH}_3\text{COOH}$) forms the salt sodium acetate ($\text{NaCH}_3\text{COO}$).

  • Carbonate ($\text{CO}_3^{2-}$) and hydrogen carbonate ($\text{HCO}_3^-$) anions from carbonic acid ($\text{H}_2\text{CO}_3$) form salts such as sodium carbonate ($\text{Na}_2\text{CO}_3$) and sodium hydrogen carbonate ($\text{NaHCO}_3$).

  • Bromide anion ($\text{Br}^-$) from hydrobromic acid ($\text{HBr}$) forms the salt sodium bromide ($\text{NaBr}$).

  • Sulfate ($\text{SO}_4^{2-}$) and hydrogen sulfate ($\text{HSO}_4^-$) anions from sulfuric acid ($\text{H}_2\text{SO}_4$) form salts such as lithium sulfate ($\text{Li}_2\text{SO}_4$) and potassium hydrogen sulfate ($\text{KHSO}_4$).

• Examples of Reactions Producing Salts:

  • Hydrobromic acid and potassium hydroxide:     $\text{HBr}(aq) + \text{KOH}(aq) \rightarrow \text{H}_2\text{O}(l) + \text{KBr}(aq)$

  • Carbonic acid and lithium hydroxide:     $\text{H}_2\text{CO}_3(aq) + \text{LiOH}(aq) \rightarrow \text{H}_2\text{O}(l) + \text{LiHCO}_3(aq)$     $\text{H}_2\text{CO}_3(aq) + 2\,\text{LiOH}(aq) \rightarrow 2\,\text{H}_2\text{O}(l) + \text{Li}_2\text{CO}_3(aq)$

  • Calcium hydroxide and phosphoric acid:     $3\,\text{Ca}(\text{OH})_2 + 2\,\text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\,\text{H}_2\text{O}$     The salt produced is calcium phosphate, $\text{Ca}_3(\text{PO}_4)_2$

  • Potassium hydroxide and nitric acid:     $\text{KOH} + \text{HNO}_3 \rightarrow \text{H}_2\text{O} + \text{KNO}_3$     The salt produced is potassium nitrate, $\text{KNO}_3$

  • Methanoic acid ($\text{HCOOH}$) and sodium hydroxide ($\text{NaOH}$):     $\text{HCOOH} + \text{NaOH} \rightarrow \text{NaHCOO} + \text{H}_2\text{O}$     The salt produced is sodium methanoate ($\text{NaHCOO}$).

  • Nitric acid ($\text{HNO}_3$) and ammonia ($\text{NH}_3$):     $\text{HNO}_3 + \text{NH}_3 \rightarrow \text{NH}_4\text{NO}_3$     The $\text{H}^+$ ion does not detach to form water; instead, $\text{NH}_3$ attaches directly to $\text{HNO}_3$ to form ammonium nitrate ($\text{NH}_4\text{NO}_3$).

  • Ethanoic acid ($\text{CH}_3\text{COOH}$) and methylamine ($\text{CH}_3\text{NH}_2$):     $\text{CH}_3\text{COOH} + \text{CH}_3\text{NH}_2 \rightarrow \text{CH}_3\text{NH}_3\text{CH}_3\text{COO}$     The acidic hydrogen from $\text{CH}_3\text{COOH}$ attaches to the amine group $\text{NH}_2$ to form methylammonium acetate ($\text{CH}_3\text{NH}_3\text{CH}_3\text{COO}$).

Acid Salts vs. Normal Salts

• Definition 1 (Structural Definition of Acid Salts): Acid salts are formed when some, but not all, of the replaceable hydrogen ions of a polyprotic acid are replaced by a metal cation or positive ion.

  • Reaction of $\text{NaOH}$ with $\text{H}_2\text{SO}_4$:     $\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O}$     $\text{NaHSO}_4$ is an acid salt because it retains an ionizable hydrogen ion ($\text{H}^+$).     $2\,\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$     $\text{Na}_2\text{SO}_4$ is a normal salt because all hydrogen ions have been replaced.

• Definition 2 (Functional/Solution-Based Definitions):

  • Acid Salts: Salts that yield an acidic solution ($\text{pH} < 7$) when dissolved in pure water.

  • Alkali Salts: Salts that yield an alkaline solution ($\text{pH} > 7$) when dissolved in pure water.

• Comparison of Definitions:

  • Lithium hydrogen carbonate ($\text{LiHCO}_3$):

  • Under Definition 1, $\text{LiHCO}_3$ is classified as an acid salt because it contains an unreplaced hydrogen ion.

  • Under Definition 2, $\text{LiHCO}_3$ is not an acid salt; it acts as an alkali salt because dissolving it in water generates a basic solution.

  • Ammonium hydrogen sulfate ($\text{NH}_5\text{SO}_4$ or $\text{NH}_4\text{HSO}_4$ formed via $\text{NH}_4\text{OH} + \text{H}_2\text{SO}_4$):

  • Under both Definition 1 and Definition 2, it is classified as an acid salt (contains replaceable hydrogen and creates an acidic solution).

• Prerequisites for Acid Salt Formation: Acid salts can only be formed from polyprotic acids, which contain more than one ionizable hydrogen atom per molecule:

  • Diprotic (dibasic) acids: $\text{H}_2\text{SO}_4$, $\text{H}_2\text{SO}_3$, $\text{H}_2\text{SO}_2$, $\text{H}_2\text{CO}_3$, $\text{H}_2\text{S}$, $\text{H}_2\text{C}_2\text{O}_4$ (oxalic acid), $\text{H}_2\text{C}_3\text{H}_2\text{O}_4$ (malonic acid).

  • Triprotic (tribasic) acids: $\text{H}_3\text{PO}_4$, $\text{H}_3\text{PO}_3$.

• Examples of Acid Salt Reactions:   $\text{H}_2\text{SO}_4 + \text{KOH} \rightarrow \text{KHSO}_4 + \text{H}_2\text{O}$   $\text{H}_2\text{CO}_3 + \text{LiOH} \rightarrow \text{LiHCO}_3 + \text{H}_2\text{O}$   $\text{H}_3\text{PO}_4 + \text{KOH} \rightarrow \text{KH}_2\text{PO}_4 + \text{H}_2\text{O}$   $\text{H}_3\text{PO}_4 + 2\,\text{KOH} \rightarrow \text{K}_2\text{HPO}_4 + 2\,\text{H}_2\text{O}$

  • Reaction between lithium hydroxide and phosphoric acid produces two distinct acid salts: potassium/lithium dihydrogen phosphate ($\text{LiH}_2\text{PO}_4$) and lithium hydrogen phosphate ($\text{Li}_2\text{HPO}_4$). The normal salt $\text{Li}_3\text{PO}_4$ can also form, but it is not an acid salt.

• Normal Salts: Normal salts are formed when all hydrogen ions in the parent acid are completely replaced by metal or ammonium cations:   $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$   $\text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O}$   $\text{H}_3\text{PO}_4 + 3\,\text{KOH} \rightarrow \text{K}_3\text{PO}_4 + 3\,\text{H}_2\text{O}$   $\text{H}_2\text{SO}_4 + 2\,\text{LiOH} \rightarrow \text{Li}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$   $\text{CH}_3\text{COOH} + \text{LiOH} \rightarrow \text{LiCH}_3\text{COO} + \text{H}_2\text{O}$

  • Potassium hydroxide reacting with sulfurous acid ($\text{H}_2\text{SO}_3$):

  • $\text{KOH} + \text{H}_2\text{SO}_3 \rightarrow \text{KHSO}_3 + \text{H}_2\text{O}$ (Acid salt: potassium hydrogen sulfite)

  • $2\,\text{KOH} + \text{H}_2\text{SO}_3 \rightarrow \text{K}_2\text{SO}_3 + 2\,\text{H}_2\text{O}$ (Normal salt: potassium sulfite)

Predicting Solution pH from Salt Hydrolysis (Qualitative Rules)

• Qualitative Rules for Dissolving Salts in Water:

  • Strong Acid + Strong Base: Produces a neutral salt ($\text{pH} = 7$).

  • Example: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

  • Strong Acid + Weak Base: Produces a salt that forms an acidic solution when dissolved in water ($\text{pH} < 7$).

  • Example: $\text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl}$

  • Weak Acid + Strong Base: Produces a salt that forms a basic solution when dissolved in water ($\text{pH} > 7$).

  • Example: $\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{NaCH}_3\text{COO} + \text{H}_2\text{O}$

  • Weak Acid + Weak Base: Produces a salt whose solution may be acidic, basic, or neutral depending on the relative dissociation constants of the parent species.

  • Example: $\text{CH}_3\text{COOH} + \text{NH}_3 \rightarrow \text{NH}_4\text{CH}_3\text{COO}$

• Chemical Mechanisms of Salt Hydrolysis:

  • Basic Solution Example ($\text{K}_2\text{CO}_3$):

  • Parent species: $\text{KOH}$ (strong base) and $\text{H}_2\text{CO}_3$ (weak acid).

  • Ionization: $\text{K}_2\text{CO}_3(aq) \rightarrow 2\,\text{K}^+ + \text{CO}_3^{2-}$

  • Hydrolysis step: Carbonate ion reacts with water:       $\text{CO}_3^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HCO}_3^- + \text{OH}^-$

  • Accumulation of $\text{OH}^-$ makes the resulting solution basic.

  • Acidic Solution Example ($\text{NH}_4\text{Cl}$):

  • Parent species: $\text{HCl}$ (strong acid) and $\text{NH}_4\text{OH}$ / $\text{NH}_3$ (weak base).

  • Ionization: $\text{NH}_4\text{Cl}(aq) \rightarrow \text{NH}_4^+ + \text{Cl}^-$

  • Hydrolysis step: Ammonium ion reacts with water:       $\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}^+$

  • Accumulation of $\text{H}^+$ makes the resulting solution acidic.

  • Neutral Solution Example ($\text{LiBr}$):

  • Parent species: $\text{HBr}$ (strong acid) and $\text{LiOH}$ (strong base).

  • Neither $\text{Li}^+$ nor $\text{Br}^-$ hydrolyzes in water ($\text{Li}^+ + \text{H}_2\text{O} \rightarrow \text{no reaction}$; $\text{Br}^- + \text{H}_2\text{O} \rightarrow \text{no reaction}$).

  • No extra $\text{H}^+$ or $\text{OH}^-$ ions are generated, yielding a neutral solution.

• Identification Heuristics for Acid/Base Components:

  • Ions deriving from strong bases: $\text{Na}^+$, $\text{Li}^+$, $\text{K}^+$, $\text{Rb}^+$ (from $\text{NaOH}$, $\text{LiOH}$, $\text{KOH}$, $\text{RbOH}$).

  • Ions deriving from strong acids: $\text{Cl}^-$, $\text{I}^-$, $\text{Br}^-$, $\text{NO}_3^-$ (from $\text{HCl}$, $\text{HI}$, $\text{HBr}$, $\text{HNO}_3$).

  • Indicators of weak acid parents: $\text{CO}_3^{2-}$, $\text{HCO}_3^-$ (from $\text{H}_2\text{CO}_3$), $\text{CH}_3\text{COOH}$.

  • Indicators of weak base parents: $\text{NH}_3$.

Weak Acid + Weak Base Salt pH Prediction (Quantitative Comparison)

• Methodology for Weak Acid + Weak Base Salts:

  • Step 1: Identify the cation's parent acid and the anion's parent base.

  • Step 2: Compare the acid ionization constant ($K_a$) of the cation against the base ionization constant ($K_b$) of the anion.

  • Step 3:

  • If $K_a > K_b$, the solution is acidic.

  • If $K_b > K_a$, the solution is basic.

  • If $K_a = K_b$, the solution is neutral.

• Worked Example 1: Ammonium Cyanide ($\text{NH}_4\text{CN}$) at $25^\circ\text{C}$:

  • Given constants: $K_a(\text{HCN}) = 6.03 \times 10^{-10}$, $K_b(\text{NH}_3) = 1.80 \times 10^{-5}$.

  • Derived constants: $K_b(\text{CN}^-) = 1.66 \times 10^{-5}$, $K_a(\text{NH}_4^+) = 5.56 \times 10^{-10}$.

  • Comparison: Since $K_b(\text{CN}^-) > K_a(\text{NH}_4^+)$ ($1.66 \times 10^{-5} > 5.56 \times 10^{-10}$), an aqueous solution of $\text{NH}_4\text{CN}$ is basic.

• Worked Example 2: Dissolving $\text{NH}_2\text{C}_6\text{H}_4\text{CH}_3\text{COO}$ in Water:

  • Identify components: Parent base is phenylamine ($\text{NH}_2\text{C}_6\text{H}_5$); parent acid is ethanoic/acetic acid ($\text{CH}_3\text{COOH}$).

  • Constants at $25^\circ\text{C}$:

  • Phenylamine ($\text{NH}_2\text{C}_6\text{H}_5$): $pK_b = 9.13 \implies K_b = 7.4 \times 10^{-10}$

  • Acetic acid ($\text{CH}_3\text{COOH}$): $pK_a = 4.76 \implies K_a = 1.7 \times 10^{-5}$

• Comparison: $K_a (1.7 \times 10^{-5}) > K_b (7.4 \times 10^{-10})$.

• Conclusion: Because $K_a > K_b$, dissolving $\text{NH}_2\text{C}_6\text{H}_4\text{CH}_3\text{COO}$ in water produces an acidic solution.

Acidity of Hydrated Metal Cations

• Behavior of Metal Cations in Water: Dissolving salts containing small, highly-charged metal ions (such as $\text{Fe}^{3+}$, $\text{Cu}^{2+}$, $\text{Al}^{3+}$) produces acidic solutions.

  • Mechanism: Small, highly-charged metal cations attract and bind $\text{OH}^-$ from water molecules, liberating free $\text{H}^+$ ions into the solution.     $\text{Fe}^{3+} + \text{H}_2\text{O} \rightarrow \text{Fe}(\text{OH})_3 + 3\,\text{H}^+$     $\text{Cu}^{2+} + \text{H}_2\text{O} \rightarrow \text{Cu}(\text{OH})_2 + 2\,\text{H}^+$     $\text{Al}^{3+} + \text{H}_2\text{O} \rightarrow \text{Al}(\text{OH})_3 + 3\,\text{H}^+$

  • Counter-Example (Large Cations): Uranium ($\text{U}^{3+}$) is a large ion despite its $+3$ charge. It does not pull $\text{OH}^-$ from water, so no reaction occurs and the solution remains neutral:     $\text{U}^{3+} + \text{H}_2\text{O} \rightarrow \text{U}^{3+} + \text{H}_2\text{O}$

• Acid Dissociation Constants ($K_a$) for Metal Cations:

• Exact values:

  • Iron(III) ($\text{Fe}^{3+}$): $K_a = 6.3 \times 10^{-3}$

  • Chromium(III) ($\text{Cr}^{3+}$): $K_a = 1.6 \times 10^{-4}$

  • Aluminum ($\text{Al}^{3+}$): $K_a = 1.4 \times 10^{-5}$

  • Iron(II) ($\text{Fe}^{2+}$): $K_a = 3.2 \times 10^{-10}$

  • Zinc ($\text{Zn}^{2+}$): $K_a = 2.5 \times 10^{-10}$

  • Nickel ($\text{Ni}^{2+}$): $K_a = 2.5 \times 10^{-11}$

• Trend: Metal cations with a $+3$ charge have $K_a$ values between $10^{-3}$ and $10^{-6}$, making them significantly more acidic than $+2$ metal cations ($K_a \approx 10^{-10}$ to $10^{-11}$).

Comprehensive Practice Problems & Applications

• Practice Problem 1: Acidic, Alkaline, or Neutral Solutions

  • Predict and explain $\text{pH}$ behavior for $0.1\,\text{mol\,dm}^{-3}$ solutions:

  • a) $0.1\,\text{mol\,dm}^{-3}\ \text{FeCl}_3(aq)$: Acidic. Contains small, highly-charged $\text{Fe}^{3+}$ ions which hydrolyze water to yield excess $\text{H}^+$.

  • b) $0.1\,\text{mol\,dm}^{-3}\ \text{NaNO}_3(aq)$: Neutral. Derived from strong base $\text{NaOH}$ and strong acid $\text{HNO}_3$; neither ion hydrolyzes.

  • c) $0.1\,\text{mol\,dm}^{-3}\ \text{Na}_2\text{CO}_3(aq)$: Basic. Derived from strong base $\text{NaOH}$ and weak acid $\text{H}_2\text{CO}_3$; $\text{CO}_3^{2-}$ hydrolyzes to produce $\text{OH}^-$.

• Practice Problem 2: Ranking Salt Solutions by Acidity

  • Task: Order $\text{NaCl}$, $\text{MgSO}_4$, $\text{Al}(\text{NO}_3)_3$, and $\text{KHCO}_3$ from most acidic to least acidic when added to water.

  • Step-by-Step Parent Analysis:

  • $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ (Strong base + Strong acid $\rightarrow$ Neutral solution)

  • $\text{Al}(\text{OH})_3 + 3\,\text{HNO}_3 \rightarrow \text{Al}(\text{NO}_3)_3 + 3\,\text{H}_2\text{O}$ (Weak base + Strong acid $\rightarrow$ Acidic solution; contains $\text{Al}^{3+}$ cation)

  • $\text{Mg}(\text{OH})_2 + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + 2\,\text{H}_2\text{O}$ (Weak base + Strong acid $\rightarrow$ Acidic solution; contains $\text{Mg}^{2+}$ cation)

  • $\text{KOH} + \text{H}_2\text{CO}_3 \rightarrow \text{KHCO}_3 + \text{H}_2\text{O}$ (Strong base + Weak acid $\rightarrow$ Basic solution)

  • Comparing $\text{Al}(\text{NO}_3)_3$ vs $\text{MgSO}_4$: $\text{Al}^{3+}$ has a higher positive charge ($+3$) than $\text{Mg}^{2+}$ ($+2$), resulting in a higher charge density and a significantly higher $K_a$. Therefore, $\text{Al}(\text{NO}_3)_3$ is more acidic than $\text{MgSO}_4$.

  • Final Order (Most Acidic to Least Acidic): $\text{Al}(\text{NO}_3)_3 > \text{MgSO}_4 > \text{NaCl} > \text{KHCO}_3$

Rigorous Calculation of Salt Solution pH

• Practice Problem: Proof of $\text{pH}$ for Sodium Acetate Solution

  • Task: Use $K_b = 5.4 \times 10^{-10}$ for the acetate ion ($\text{CH}_3\text{COO}^-$) at $25^\circ\text{C}$ to prove that the $\text{pH}$ of a $0.10\,\text{mole/L}$ solution of $\text{NaCH}_3\text{COO}$ is approximately 8.9.

• Step 1: Dissociation and Equilibrium Equations

  • Dissociation in water:     $\text{NaCH}_3\text{COO}(aq) \rightarrow \text{Na}^+(aq) + \text{CH}_3\text{COO}^-(aq)$

  • Base hydrolysis reaction:     $\text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq)$

  • Equilibrium expression:     $K_b = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]}$…

• Step 2: Setting up ICE Values and Substitution

  • Initial concentration of CH3COO−=0.10 mol dm−3CH3​COO−=0.10moldm−3

  • Let x=[CH3COOH]=[OH−]x=[CH3​COOH]=[OH−] at equilibrium.

  • Equilibrium concentration of CH3COO−=0.10−xCH3​COO−=0.10−x

  • Substitute into the KbKb​ expression:

    Kb=x⋅x0.10−x=x20.10−xKb​=0.10−xx⋅x​=0.10−xx2​

    5.4×10−10=x20.10−x5.4×10−10=0.10−xx2​

• Step 3: Solving for xx and Equilibrium Concentrations

  • Solving the quadratic equation yields:

    x=7.3×10−6x=7.3×10−6

• Resulting equilibrium concentrations:

  • [OH−]=7.3×10−6 mol dm−3[OH−]=7.3×10−6moldm−3

  • [CH3COOH]=7.3×10−6 mol dm−3[CH3​COOH]=7.3×10−6moldm−3

  • [CH3COO−]=0.1−7.3×10−6=0.0999993 mol dm−3[CH3​COO−]=0.1−7.3×10−6=0.0999993moldm−3

  • Step 4: Incorporating Water Auto-Ionization and Calculating pHpH

• Water auto-ionization constant at 25∘C25∘C: Kw=1.008×10−14Kw​=1.008×10−14

• Kw=[H+][OH−]Kw​=[H+][OH−]

• Setting up system considering total [OH−][OH−]:

    1.008×10−14=[H+](7.3×10−6)1.008×10−14=[H+](7.3×10−6)

    [H+]=1.008×10−147.3×10−6=1.36961×10−9 mol dm−3[H+]=7.3×10−61.008×10−14​=1.36961×10−9moldm−3

• Calculating pHpH:

    pH=−log⁡10(1.36961×10−9)=8.86pH=−log10​(1.36961×10−9)=8.86

• Concluding result: The pHpH is approximately 8.98.9

Titration Equivalence Points and Indicator Selection

• Titration Overview: An acid-base titration is an experimental procedure in which measured volumes of an acid/base are added to a base/acid to reach complete neutralization.

• Equivalence Point Definition: The equivalence point is the exact moment in a titration when the number of moles of H+H+ added equals the number of moles of OH−OH− added ([H+]=[OH−][H+]=[OH−]).

  • Example molar ratios:

  • 100 cm3100cm3 of 1.000 mol dm−3 HCl1.000moldm−3 HCl + 100 cm3100cm3 of 1.000 mol dm−3 NaOH1.000moldm−3 NaOH: Solution is at the equivalence point ([H+]=[OH−][H+]=[OH−]).

  • 100 cm3100cm3 of 1.000 mol dm−3 HCl1.000moldm−3 HCl + 90 cm390cm3 of 1.000 mol dm−3 NaOH1.000moldm−3 NaOH: Solution is before the equivalence point (excess H+H+).

  • 100 cm3100cm3 of 1.000 mol dm−3 HCl1.000moldm−3 HCl + 110 cm3110cm3 of 1.000 mol dm−3 NaOH1.000moldm−3 NaOH: Solution is past the equivalence point (excess OH−OH−).

• Equivalence Point pHpH Varies by Acid-Base Strength:

  • Strong Acid + Strong Base Titration: Forms a neutral salt. Solution pH=7pH=7 at the equivalence point.

  • Weak Acid + Strong Base Titration: Forms a basic salt. Solution pH>7pH>7 at the equivalence point.

  • Strong Acid + Weak Base Titration: Forms an acidic salt. Solution pH<7pH<7 at the equivalence point.

• Role and Limitations of Phenolphthalein:

  • Phenolphthalein is a chemical indicator that transitions from colorless to purple/pink when pHpH rises above 77.

  • Inapplicability to Strong Acid + Weak Base Titrations:

  • In the titration HCl+NH3→NH4ClHCl+NH3​→NH4​Cl, the salt NH4ClNH4​Cl makes the solution acidic at the equivalence point (pH<7pH<7).

  • If phenolphthalein is used, the color change occurs only after adding excess base to bring the pHpH above 77. This causes a significant overestimation of the base required to reach equivalence.

  • Rule for Phenolphthalein Use: Phenolphthalein is suitable only when the salt formed at the equivalence point is neutral or basic.

• Indicator Selection Rule: Indicators must be selected such that their color transition pHpH range corresponds to the pHpH at the titration's equivalence point.

  • Practice Question: What is an appropriate indicator for a titration that produces a salt with an equivalence point $$\text