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⏱️General Chemistry II Unit 3 Review

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3.3 pH scale and calculations for strong and weak acids/bases

⏱️General Chemistry II
Unit 3 Review

3.3 pH scale and calculations for strong and weak acids/bases

Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
⏱️General Chemistry II
Unit & Topic Study Guides

pH is a crucial concept in chemistry, measuring the acidity or basicity of solutions. It's defined as the negative logarithm of hydronium ion concentration, ranging from 0 to 14. Understanding pH helps us grasp how acids and bases behave in various contexts.

Calculating pH for strong and weak acids and bases involves different approaches. Strong acids and bases completely dissociate, making pH calculations straightforward. Weak acids and bases partially dissociate, requiring the use of dissociation constants and equilibrium calculations to determine pH.

pH and the Concentration of Hydronium Ions

Definition and significance of pH

  • pH logarithmic scale measures acidity or basicity of a solution
    • Ranges from 0 to 14 with 7 being neutral (pure water at 25℃)
    • Lower pH values indicate higher acidity (lemon juice, vinegar)
    • Higher pH values indicate higher basicity (baking soda, bleach)
  • pH defined as negative logarithm base 10 of hydronium ion concentration $[H_3O^+]$ in a solution
    • Mathematical expression: $pH = -\log_{10}[H_3O^+]$
  • Concentration of hydronium ions $[H_3O^+]$ inversely related to pH
    • As $[H_3O^+]$ increases, pH decreases and solution becomes more acidic
    • As $[H_3O^+]$ decreases, pH increases and solution becomes more basic
  • In pure water at 25℃, $[H_3O^+] = [OH^-] = 1 \times 10^{-7}$ M resulting in neutral pH of 7
    • Acidic solutions have $[H_3O^+] > 1 \times 10^{-7}$ M and $pH < 7$
    • Basic solutions have $[H_3O^+] < 1 \times 10^{-7}$ M and $pH > 7$

Calculating pH for Strong and Weak Acids and Bases

pH calculations for strong acids and bases

  • Strong acids and bases completely dissociate in aqueous solutions
    • For strong acid HA: $HA_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+{(aq)} + A^-{(aq)}$ (hydrochloric acid HCl, nitric acid HNO₃)
    • For strong base MOH: $MOH_{(aq)} \rightarrow M^+{(aq)} + OH^-{(aq)}$ (sodium hydroxide NaOH, potassium hydroxide KOH)
  • For strong acids, $[H_3O^+]$ equals initial concentration of acid
    • Calculate pH using: $pH = -\log_{10}[H_3O^+]$
  • For strong bases, $[OH^-]$ equals initial concentration of base
    • Calculate pOH using: $pOH = -\log_{10}[OH^-]$
    • Convert pOH to pH using: $pH + pOH = 14$

pH determination for weak acids and bases

  • Weak acids and bases partially dissociate in aqueous solutions
    • For weak acid HA: $HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+{(aq)} + A^-{(aq)}$ (acetic acid CH₃COOH, benzoic acid C₆H₅COOH)
    • For weak base B: $B_{(aq)} + H_2O_{(l)} \rightleftharpoons BH^+{(aq)} + OH^-{(aq)}$ (ammonia NH₃, methylamine CH₃NH₂)
  • Acid dissociation constant Ka used to determine pH of weak acids
    • $K_a = \frac{[H_3O^+][A^-]}{[HA]}$
    • Assume $[H_3O^+] = [A^-]$ and $[HA]\text{initial} \approx [HA]\text{equilibrium}$
    • Solve for $[H_3O^+]$ using quadratic formula or approximation methods
    • Calculate pH using: $pH = -\log_{10}[H_3O^+]$
  • Base dissociation constant Kb used to determine pH of weak bases
    • $K_b = \frac{[BH^+][OH^-]}{[B]}$
    • Assume $[BH^+] = [OH^-]$ and $[B]\text{initial} \approx [B]\text{equilibrium}$
    • Solve for $[OH^-]$ using quadratic formula or approximation methods
    • Calculate pOH using: $pOH = -\log_{10}[OH^-]$
    • Convert pOH to pH using: $pH + pOH = 14$

Converting Between pH, pOH, [H+], and [OH-]

  • pH and pOH related by equation: $pH + pOH = 14$
    • If given pH, calculate pOH using: $pOH = 14 - pH$
    • If given pOH, calculate pH using: $pH = 14 - pOH$
  • $[H_3O^+]$ and $[OH^-]$ related by ion product of water, $K_w = [H_3O^+][OH^-] = 1 \times 10^{-14}$ at 25℃
    1. If given $[H_3O^+]$, calculate $[OH^-]$ using: $[OH^-] = \frac{K_w}{[H_3O^+]}$
    2. If given $[OH^-]$, calculate $[H_3O^+]$ using: $[H_3O^+] = \frac{K_w}{[OH^-]}$
  • To convert between pH and $[H_3O^+]$, use:
    • $pH = -\log_{10}[H_3O^+]$
    • $[H_3O^+] = 10^{-pH}$
  • To convert between pOH and $[OH^-]$, use:
    • $pOH = -\log_{10}[OH^-]$
    • $[OH^-] = 10^{-pOH}$