Acids, Bases, pH, and Dissociation Constants

Acids and Bases

Acids

  • Substances that dissociate in water to give H+ (Hydrogen ion) or H3O+H_3O^+ (Hydronium ion).
  • Strong acids: Ionize completely in water.
    • Example: HCl(aq)→H+(aq)+Cl−(aq)HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)
    • Example: H<em>2SO</em>4(aq)→2H+(aq)+SO42−(aq)H<em>2SO</em>4(aq) \rightarrow 2H^+(aq) + SO_4^{2-}(aq)

Bases

  • Substances that dissociate in water to give OH−OH^- (Hydroxide ion).
  • Strong bases: Ionize completely in water.
    • Example: NaOH(aq)→Na+(aq)+OH−(aq)NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)
    • Example: Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)Ca(OH)_2(aq) \rightarrow Ca^{2+}(aq) + 2OH^-(aq)

Weak Acids and Bases

  • Ionize partially in water.
    • Example: H<em>3PO</em>4⇌3H++PO43−H<em>3PO</em>4 \rightleftharpoons 3H^+ + PO_4^{3-}
  • Equilibrium arrows indicate that a reaction can proceed in both directions (reversible).
  • Acids and bases can be strong (completely ionized) or weak (partially ionized).

pH Scale

Definition

  • A logarithmic scale used to specify how acidic or basic a solution is.
  • pH scale ranges from 0 to 14.
    • Acids: pH from 0 to 6.
    • Neutral: pH = 7.
    • Bases: pH from 8 to 14.

pH Ranges

  • Strong acids: pH from 1 to 3.
  • Weak acids: pH from 3 to 6.
  • Strong bases: pH from 12 to 14.
  • Weak bases: pH from 8 to 11.

Salts

  • pH from 8.49 to 8.19. (This seems like an outlier/error and needs checking).

Logarithmic Scale

  • Illustrates the difference in pH values between strong and weak acids/bases.

pH, pOH, and Ion Concentrations

Formulas

  • pH=−log[H+]pH = -log[H^+]
  • [H+]=10−pH[H^+] = 10^{-pH}
  • pOH=−log[OH−]pOH = -log[OH^-]
  • [OH−]=10−pOH[OH^-] = 10^{-pOH}
  • [H+][H^+] represents the hydrogen (or hydronium) ion concentration.
  • [OH−][OH^-] represents the hydroxide ion concentration.

Examples

  • 0.001 M HCl:
    • pH=−log[0.001]=3pH = -log[0.001] = 3
  • 0.004 M Ca(OH)2Ca(OH)_2:
    • pOH=−log[0.008]pOH = -log[0.008] (Note: 0.004 M Ca(OH)2Ca(OH)_2 gives 0.008 M OH−OH^-, since it dissociates into 2 OH−OH^- ions)
    • pOH=2.10pOH = 2.10

Calculating Hydrogen Ion Concentration from pH

  • If the pH of a solution is 3.5, then:
    • [H+]=10−3.5=0.00032M[H^+] = 10^{-3.5} = 0.00032 M

Relationship Between pH and pOH

Equation

  • Based on equilibrium concentrations of H+H^+ and OH−OH^- in water at 25°C:
    • pH+pOH=14pH + pOH = 14

Example 1

  • What is the pH of a 6.2×10−5M6.2 \times 10^{-5} M NaOH solution?
    • pOH=−log[6.2×10−5]pOH = -log[6.2 \times 10^{-5}]
    • pOH=4.2pOH = 4.2
    • pH=14−4.2=9.8pH = 14 - 4.2 = 9.8

Example 2

  • A solution is created by measuring 3.60×10−33.60 \times 10^{-3} moles of NaOH and 5.95×10−45.95 \times 10^{-4} moles of HCl into a container, and then water is added until the final volume is 1.00 L. What is the pH of this solution?
    • Moles of NaOH remaining after neutralization: 3.60×10−3−5.95×10−4=3.005×10−33.60 \times 10^{-3} - 5.95 \times 10^{-4} = 3.005 \times 10^{-3} moles
    • Molarity of NaOH: 3.005×10−3M3.005 \times 10^{-3} M
    • pOH=−log[3.005×10−3]pOH = -log[3.005 \times 10^{-3}]
    • pOH=2.522pOH = 2.522
    • pH=14−2.522=11.48pH = 14 - 2.522 = 11.48

Acid and Base Dissociation Constants

Ka (Acid Dissociation Constant)

  • Quantifies the strength of acids, reflecting how readily they donate H+H^+.
  • Higher KaK_a means a stronger acid.
  • For a weak acid HA, the reaction is: HA+H<em>2O⇌H</em>3O++A−HA + H<em>2O \rightleftharpoons H</em>3O^+ + A^-
    • K<em>a=[H</em>3O+][A−][HA]K<em>a = \frac{[H</em>3O^+][A^-]}{[HA]}

Kb (Base Dissociation Constant)

  • Quantifies the strength of bases, reflecting how readily they accept H+H^+ or donate electrons.
  • Higher KbK_b means a stronger base.
  • For a weak base B, the reaction is: B+H2O⇌BH++OH−B + H_2O \rightleftharpoons BH^+ + OH^-
    • Kb=[BH+][OH−][B]K_b = \frac{[BH^+][OH^-]}{[B]}

Relationship Between Ka and Kb

  • For a conjugate acid-base pair:
    • K<em>a×K</em>b=KwK<em>a \times K</em>b = K_w
    • KwK_w (ion product of water) is approximately 1.0×10−141.0 \times 10^{-14} at 25°C.

Acid-Base Conjugate Pairs

  • The stronger the acid, the weaker its conjugate base, and vice versa.
  • Acid donates H+H^+, conjugate base accepts H+H^+.

pKa and pKb

Definitions

  • pK<em>apK<em>a is the negative logarithm of K</em>aK</em>a, used to compare the acidity of different acids.
    • pK<em>a=−log[K</em>a]pK<em>a = -log[K</em>a]
    • K<em>a=10−pK</em>aK<em>a = 10^{-pK</em>a}
  • pK<em>bpK<em>b is the negative logarithm of K</em>bK</em>b, used to measure the degree of dissociation for a base.
    • pK<em>b=−log[K</em>b]pK<em>b = -log[K</em>b]
    • K<em>b=10−pK</em>bK<em>b = 10^{-pK</em>b}

Relationship

  • pK<em>a+pK</em>b=pKwpK<em>a + pK</em>b = pK_w

Examples

  • What is the pK<em>bpK<em>b for methylamine if the value of K</em>bK</em>b for methylamine is 4.4×10−44.4 \times 10^{-4}?
    • pKb=−log[4.4×10−4]=3.36pK_b = -log[4.4 \times 10^{-4}] = 3.36
  • For NH<em>3NH<em>3, K</em>b=1.8×10−5K</em>b = 1.8 \times 10^{-5}:
    • Write the formula for the conjugate acid: NH4+NH_4^+
    • Find KaK_a for the conjugate acid:
      • K<em>a=K</em>wKb=1.0×10−141.8×10−5=5.6×10−10K<em>a = \frac{K</em>w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}