Acids-and-Bases-1

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Acids and Bases


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Brønsted-Lowry Acids and Bases

  • The Brønsted-Lowry Theory explains acid-base reactions through proton transfer.
  • A hydrogen atom (H) contains one proton and one electron, therefore a proton (H+) is synonymous with a hydrogen ion.
  • Brønsted-Lowry Acid: A proton donor that releases hydrogen ions (H+) in water forming hydroxonium ions (H3O+).
    • Reaction Example: HA(aq) + H2O → H3O+(aq) + A+(aq)
  • Brønsted-Lowry Base: A proton acceptor that takes hydrogen ions (H+) from water.
    • Reaction Example: B(aq) + H2O → BH+(aq) + OH-(aq)

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Conjugate Acid-Base Pairs

  • Brønsted-Lowry Theory defines acids and bases as conjugate pairs, capable of transforming into each other through proton transfer.
  • Acid-base reactions are equilibria.
    • General Reaction: HA ⇌ H+ + A-
  • In the reaction, HA donates a proton and transforms into its conjugate base A-, whereas the conjugate base A- accepts a proton to form back the conjugate acid HA.

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Conjugate Acid-Base Examples

  1. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
  2. HCO3-(aq) + S2-(aq) ⇌ HS-(aq) + CO32-(aq)
  • Indication of acid (HA) and base (B) along with their conjugate pairs.

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Strong Acids and Bases

  • Strong Acids: Dissociate almost completely in water and release most H+.
    • Example: HCl → H+ + Cl-
    • Other strong acids include HNO3 (nitric acid), H2SO4 (sulfuric acid).
  • Strong Bases: Also dissociate almost completely in water.
    • Example: NaOH → Na+ + OH-

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Weak Acids and Bases

  • Most acids and bases are weak, partially ionizing in water and maintaining an equilibrium.
  • Weak Acids: Slightly dissociate in water with a shift in equilibrium to the left.
    • Example: CH3COOH ⇌ CH3COO- + H+
  • Weak Bases: Also slightly dissociate in water.
    • Example: NH3 + H2O ⇌ NH4+ + OH-
  • Only about 2% of ethanoic acid (CH3COOH) dissociates in solution.

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Acid-Base Equilibria

  • An acid requires a base to accept a proton.
  • Proton transfer occurs between acids (HA) and bases (B).
    • Reaction: HA(aq) + B(aq) ⇌ BH+(aq) + A-(aq)
  • Changes in concentrations of acid or base will shift the equilibrium to maintain balance:
    • Adding HA or B shifts equilibrium right.
    • Adding BH+ or A- shifts equilibrium left.

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Dissociation of Water

  • Water can also dissociate slightly, creating an equilibrium:
    • Reaction: H2O + H2O ⇌ H3O+ + OH-
    • Alternative representation: H2O ⇌ H+ + OH-
  • The equilibrium constant expression can be determined, showing very low concentrations of ions lead to constant water concentrations.

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Kw

  • The ionic product of water (Kw) is a constant, given by:
    • Kw = 1.0 x 10^-14 mol² dm⁻⁶ at 298 K.
  • The value of Kw varies with temperature, and in pure water, [H+] = [OH-]. Thus, Kw = [H+]².

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Determination of pH

  • pH is defined as the negative logarithm of the molar hydrogen-ion concentration:
    • Formula: pH = -log10[H+]
  • The pH scale ranges from 0 (very acidic) to 14 (very basic), with 7 being neutral.
  • Example Calculation: For a 0.005 mol dm⁻³ hydrogen ion concentration:
    • pH = -log10(0.005) = 2.3

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Calculating [H+]

  • Hydrogen ion concentration can be derived from pH:
    • Formula: [H+] = 10^-pH.
  • Example Calculation: Given a solution of hydrochloric acid with a pH of 2.0:
    • [H+] = 10^-2.0 = 0.01 mol dm⁻³.

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Monoprotic Acids

  • Strong acids ionize fully in solution; examples include HCl and HNO3.
  • Monoprotic acids release one proton when dissociated:
    • One mole of acid yields one mole of hydrogen ions.
    • Hydrogen ion concentration equals acid concentration.
  • Example Calculation: For a 0.05 mol dm⁻³ HCl:
    • pH = -log10(0.05) = 1.3.

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Diprotic Acids

  • Strong diprotic acids release two protons per dissociation.
  • Each mole of diprotic acid yields two moles of hydrogen ions.
  • Example Calculation: For a 0.01 mol dm⁻³ sulfuric acid (H2SO4):
    • [H2SO4] = 0.01 mol dm⁻³; therefore,
    • [H+] = 2 × 0.01 mol dm⁻³ = 0.02 mol dm⁻³;
    • pH = -log10(0.02) = 1.7.

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Calculating pH of Strong Base

  • Strong bases fully ionize in water (e.g., NaOH → Na+ + OH-).
  • Concentrations of bases align with hydroxide ion concentrations:
    • For a strong base, [Strong base] = [OH-].
  • To find pH, calculate [H+] using Kw:
    • Kw = [H+][OH-].

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Base pH Calculation Example

  • Example: Calculate the pH of 0.20 mol dm⁻³ NaOH at 298 K:
    • Given Kw = 1.0 x 10^-14, use Kw equation:
    • 1.0 x 10^-14 = [H+] × 0.20 mol dm⁻³
    • Rearrange:
    • [H+] = 1.0 x 10^-14 / 0.20 = 5 x 10^-14 mol dm⁻³;
    • pH = -log10[H+] = -log10(5 x 10^-14) = 13.3.

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Ka

  • Ka is the acid dissociation constant; it requires two assumptions:
    • Weak acids dissociate slightly in solutions; [H+] is not equal to [acid].
    • Use Ka to find pH:
    • Dissociation: HA(aq) ⇌ H+(aq) + A-(aq).
    • Assume equilibrium concentration of HA = initial concentration of HA.
    • Assumption: [H+] = [A-].

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Using Ka to find pH

  • To calculate the pH of a weak acid using Ka:
    • Ka is a specific constant for particular acid at fixed temperature.
  • Example: For 0.030 mol dm⁻³ hydrogen fluoride at 298 K with Ka = 6.6 x 10^-4:
    • The Ka expression: Ka = [H+]² / [HA].

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Using Ka to find pH

  • Rearranging the Ka expression to find [H+]:
    • [H+] = √(Ka × [HA])
  • Calculation:
    • [H+] = √(6.6 x 10^-4 × 0.030) = 1.98 x 10^-5.
  • Find pH:
    • pH = -log10[H+] = -log10(1.98 x 10^-5) = 4.70.

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Using Ka to find Concentration

  • Ka allows the calculation of the concentration of a weak acid when given pH and Ka.
  • Example: Calculate molar concentration of propanoic acid solution with pH 3.30 and Ka = 1.30 x 10^-5:
    • First calculate [H+]: [H+] = 10^-pH = 10^-3.30 = 5.01 x 10^-4.

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Using Ka to find Concentration

  • Write the Ka expression to find [CH3CH2COOH]:
    • Ka = [H+]² / [CH3CH2COOH], rearranging gives [CH3CH2COOH] = [H+]² / Ka.
  • Substitute into equation:
    • [CH3CH2COOH] = (5.01 x 10^-4)² / (1.30 x 10^-5) = 0.0193 mol dm⁻³.

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pKa

  • pKa indicates acid strength:
    • Lower pKa values indicate stronger acids.
  • Conversion equation:
    • pKa = -log10(Ka);
    • Alternatively, Ka = 10^-pKa.

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pKa Calculations

  • Example for propanoic acid with Ka = 1.30 x 10^-5:
    • Calculate pKa:
    • pKa = -log10(1.30 x 10^-5) = 4.89.
  • Example for HF with pKa 3.18 to find Ka:
    • Ka = 10^-3.18 = 6.60 x 10^-4.
    • Frequently, concentration calculations from pH may require conversion of pKa to Ka first.

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Titration Overview

  • Description of acid-base titration process:
    • Unknown acid concentration in a conical flask, while base with known concentration added from a burette.
    • Use of indicator to signal neutralization of acid by base.
    • The amount of base used provides the means to calculate the acid's concentration.

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Titrations Best Practices

  • To achieve precision in titration results:
    • Measure the unknown volume accurately.
    • Repeat the titration at least three times for mean calculation.
    • Ensure all results are within 0.1 cm³ of each other, disregarding any anomalies.

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pH Curves

  • Visualization of pH changes as bases titrate with acids.
  • Key curves to note:
    • Strong acid - strong base
    • Strong acid - weak base
    • Weak acid - strong base
    • Weak acid - weak base

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pH Curves Analysis

  • Strong Acid - Strong Base:
    • Starts at pH of 13 before acid addition; drops to 1.
  • Weak Acid - Strong Base:
    • Begins at pH of 13, with a decrease to a pH around 5 after adding weak acid.
  • Strong Acid - Weak Base:
    • Initiates from the weak base's pH (around 9), dropping to strong acid pH (around 1).
  • Weak Acid - Weak Base:
    • Begins at weak base's pH (around 9), moving towards the weak acid's pH (around 5).

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pH Curve Details

  • Each graph (except for weak acid - weak base) exhibits a vertical section indicating neutralization ends.
  • A small addition of acid causes drastic pH changes, whereas weak acid - weak base shows gradual shifts.
  • pH meters are preferred for determining endpoints over color-change indicators.

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Indicators in Titration

  • An indicator changes color to indicate titration endpoints; it must occur precisely at the end point.
  • The selected indicator should transition within a narrow pH range, located at the steep part of the pH curve.

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Common Indicators Used

  • Strong Acid – Strong Base: Methyl orange and Phenolphthalein can be used.
  • Strong Acid – Weak Base: Methyl Orange is preferred.
  • Weak Acid – Strong Base: Use Phenolphthalein.
  • Weak Acid – Weak Base: Neither indicator is suitable due to gradual pH changes.
IndicatorColor at Low pHpH of Color ChangeColor at High pH
Methyl OrangeRED3.1 – 4.4YELLOW
PhenolphthaleinCOLOURLESS8.3 – 10PINK

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Titration Calculations

  • The results from acid/base titrations yield unknown concentration:
    • Example: 55 cm³ of HCl neutralized by 30 cm³ of 0.8 mol dm⁻³ NaOH.
    • Balanced equation: NaOH + HCl → NaCl + H2O.
  • Calculating moles of NaOH:
    • Moles = (0.8 x 30) / 1000 = 0.024 mol.

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Further Calculations on Titration

  • Reaction: NaOH + HCl → NaCl + H2O.
  • Since 1 mol of NaOH neutralizes 1 mol of HCl:
    • 0.024 mol of NaOH equates with 0.024 mol of HCl.
  • Finding HCl concentration:
    • Concentration = (0.024 x 1000) / 55 = 0.44 mol dm⁻³.

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Titration Calculations with Diprotic Acids

  • In titrations involving diprotic acids:
    • The volume vs. pH curve shows reactions completing at two stages.

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Example of Diprotic Acid Calculation

  • Example: 25 cm³ of 0.35 mol dm⁻³ H2SO4 neutralized by 30 cm³ of KOH.
    • Write balanced equation: 2KOH + H2SO4 → K2SO4 + 2H2O.
  • Calculate moles of H2SO4:
    • Moles = (0.35 x 25) / 1000 = 0.0088 mol.

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Further Diprotic Acid Concentration Calculations

  • For equation 2KOH + H2SO4 → K2SO4 + 2H2O:
    • 1 mole of H2SO4 neutralizes 2 moles of KOH;
    • 0.0088 mol of H2SO4 neutralizes 0.0176 mol of KOH.
  • Concentration of KOH calculation:
    • Concentration = (0.0176 x 1000) / 30 = 0.58 mol dm⁻³.

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Buffers

  • Buffers resist changes in pH when small amounts of acid or base are added or when diluted.
    • Even a tiny amount of acid can drastically adjust pH in water.
    • Buffers diminish large pH fluctuations but do not completely avoid pH changes.
    • Both acidic and basic buffers are applicable.

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Acidic Buffers

  • Acidic buffer solutions comprise a weak acid and its salt; they possess a pH <7.
    • Example: Ethanoic acid (CH3COOH) and Sodium ethanoate (CH3COO-Na+).
  • The solution consists of undissociated ethanoic acid and dissociated ethanoate ions.
    • Adding acid (H+) increases concentration; equilibrium shifts left to counteract.

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Acidic Buffer Responses

  • For small acid additions, excess H+ react with CH3COO- to form CH3COOH, stabilizing pH.
  • For small base additions, OH- reacts with H+ to form water, shifting equilibrium right to produce more H+ until pH stabilizes.

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Basic Buffers

  • Basic buffer solutions consist of a weak base and the salt of that weak base; they have a pH >7.
    • Example: Ammonia (NH3) and Ammonium chloride (NH4Cl).
  • The solution contains undissociated ammonia and ammonium ions from dissociated salt.
    • Adding acid (H+) reduces OH- concentration; equilibrium shifts right to maintain pH.

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Basic Buffers Response

  • Acid addition increases H+, which forms water with OH-, reducing OH- concentration and shifting equilibrium to create more OH- until returning to original pH.
  • Base addition raises OH- levels, reacting with NH4+ to yield NH3 and H2O, thus shifting equilibrium left to neutralize excess OH-.

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Calculating Buffer pH

  • To compute pH of a buffer, knowledge of Ka and concentrations of weak acid and salt is necessary:
    • Example: pH of buffer containing 0.30 mol dm⁻³ ethanoic acid and 0.50 mol dm⁻³ sodium ethanoate, with Ka = 1.7 x 10^-5.
  • Ka expression for the buffer:
    • CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq).

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Buffer pH Calculation Steps

  • Rearranging the Ka expression to solve for [H+]:
    • [H+] = Ka × [CH3COOH] / [CH3COO-].
  • Insert known values:
    • [H+] = (1.7 x 10^-5) × (0.30) / (0.50) = 1.02 x 10^-5.
  • Find pH from [H+]:
    • pH = -log10(1.02 x 10^-5) = 4.92.

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Buffer Applications

  • Biological buffers are crucial to maintain blood pH near 7.4 to protect cells and organs.
  • Hair care products often utilize buffers to prevent damage via alkaline shampoos, maintaining a pH of about 5.5.
  • Biological washing powders integrate buffers to optimize enzyme functionality for effective cleaning.