pH Calculations: Strong and Weak Acids & Bases

pH Definition

• pH (potential of Hydrogen) is the measure of H^+ concentration.

pH Equations

• Equation for pH:
  pH = -log[H^+]
• Rearranged equation to find [H^+]:
  [H^+] = log^{-1}(-pH)
• pOH is the measure of the OH^- concentration.
• Equation for pOH:
  pOH = -log[OH^-]
• Rearranged equation to find [OH^-]:
  [OH^-] = log^{-1}(-pOH)
• Relationship between pH and pOH:
  pH + pOH = 14

pH of Strong Acids

• Example: Calculate the pH of a 0.01M solution of HCl.
  HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)
• Note: Strong acids have complete dissociation, thus, [HCl] = [H^+].
• Calculation:
  pH = -log[H^+] = -log(0.01M) = 2

pH of Strong Bases

• Example: Calculate the pH of a 0.01M solution of NaOH.
  NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)
• Note: Strong bases have complete dissociation, so [NaOH] = [OH^-].
• Calculation of pOH:
  pOH = -log[OH^-] = -log(0.01) = 2
• Calculation of pH:
  pH = 14 - pOH = 14 - 2 = 12

Alternative Method for Strong Bases

• Water dissociation constant:
  K_w = [H^+][OH^-] = 1.0 \times 10^{-14}
• Given [OH^-] = 0.01M, calculate [H^+].
  [H^+] = \frac{1.0 \times 10^{-14}}{0.01} = 1.0 \times 10^{-12}
• Calculate pH:
  pH = -log[H^+] = -log(1.0 \times 10^{-12}) = 12

pH of Weak Acids

• Example: Calculate the pH of a 0.6M solution of HF (K_a = 7.0 \times 10^{-14}).
  HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)


• Note: Weak acids do not completely ionize, so [HF] \neq [H^+].


• ICE (Initial, Change, Equilibrium) table:
  

• Write the expression for the acid dissociation constant:
  K_a = \frac{[H^+(aq)][F^-(aq)]}{[HF(aq)]} = \frac{(x)(x)}{(0.60-x)}
• Assumption: x is very small, so 0.60 - x ≈ 0.60
  7.0 \times 10^{-14} = \frac{x^2}{0.60}
• Solve for x:
  x = \sqrt{(7.0 \times 10^{-14}) \times (0.60)} = 2.04 \times 10^{-7} M
• Calculate pH:
  pH = -log(2.04 \times 10^{-7}) = 6.69

pH of Weak Bases

• Example: Calculate the pH of a 0.4M solution of NH3 (Kb = 1.8 \times 10^{-5}).
  NH3(aq) (in H2O) \rightleftharpoons NH_4^+(aq) + OH^-(aq)
• Note: Weak bases do not completely ionize, so [NH_3] \neq [OH^-].
• ICE table:
  | | NH3 | NH4^+ | OH- |
  |-------------|----------|------------|------|
  | Initial | 0.40 | 0 | 0 |
  | Change | -x | +x | +x |
  | Final | 0.40-x | x | x |

Weak Base Calculations

• Write the expression for the base dissociation constant:
  Kb = \frac{[NH4^+(aq)][OH^-(aq)]}{[NH_3(aq)]} = \frac{(x)(x)}{(0.40-x)}
• Assumption: x is very small, so 0.40 - x ≈ 0.40
  1.8 \times 10^{-5} = \frac{x^2}{0.40}
• Solve for x:
  x = \sqrt{(1.8 \times 10^{-5}) \times (0.40)} = 2.68 \times 10^{-3} M
• Calculate pOH:
  pOH = -log(2.68 \times 10^{-3}) = 2.57
• Calculate pH:
  pH = 14 - 2.57 = 11.43