pH Calculations: Strong and Weak Acids & Bases

pH Definition

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

pH Equations

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

pH of Strong Acids

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

pH of Strong Bases

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

Alternative Method for Strong Bases

  • Water dissociation constant:
    Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0 \times 10^{-14}
  • Given [OH−]=0.01M[OH^-] = 0.01M, calculate [H+][H^+].
    [H+]=1.0×10−140.01=1.0×10−12[H^+] = \frac{1.0 \times 10^{-14}}{0.01} = 1.0 \times 10^{-12}
  • Calculate pH:
    pH=−log[H+]=−log(1.0×10−12)=12pH = -log[H^+] = -log(1.0 \times 10^{-12}) = 12

pH of Weak Acids


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

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

  • ICE (Initial, Change, Equilibrium) table:

HFH+F-
Initial0.6000
Change-x+x+x
Final0.60-xxx

Weak Acid Calculations

  • Write the expression for the acid dissociation constant:
    Ka=[H+(aq)][F−(aq)][HF(aq)]=(x)(x)(0.60−x)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
    7.0×10−14=x20.607.0 \times 10^{-14} = \frac{x^2}{0.60}
  • Solve for x:
    x=(7.0×10−14)×(0.60)=2.04×10−7Mx = \sqrt{(7.0 \times 10^{-14}) \times (0.60)} = 2.04 \times 10^{-7} M
  • Calculate pH:
    pH=−log(2.04×10−7)=6.69pH = -log(2.04 \times 10^{-7}) = 6.69

pH of Weak Bases

  • Example: Calculate the pH of a 0.4M solution of NH<em>3NH<em>3 (K</em>b=1.8×10−5K</em>b = 1.8 \times 10^{-5}).
    NH<em>3(aq)(inH</em>2O)⇌NH4+(aq)+OH−(aq)NH<em>3(aq) (in H</em>2O) \rightleftharpoons NH_4^+(aq) + OH^-(aq)
  • Note: Weak bases do not completely ionize, so [NH3]≠[OH−][NH_3] \neq [OH^-].
  • ICE table:
    | | NH<em>3NH<em>3 | NH</em>4+NH</em>4^+ | 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:
    K<em>b=[NH</em>4+(aq)][OH−(aq)][NH3(aq)]=(x)(x)(0.40−x)K<em>b = \frac{[NH</em>4^+(aq)][OH^-(aq)]}{[NH_3(aq)]} = \frac{(x)(x)}{(0.40-x)}
  • Assumption: x is very small, so 0.40 - x ≈ 0.40
    1.8×10−5=x20.401.8 \times 10^{-5} = \frac{x^2}{0.40}
  • Solve for x:
    x=(1.8×10−5)×(0.40)=2.68×10−3Mx = \sqrt{(1.8 \times 10^{-5}) \times (0.40)} = 2.68 \times 10^{-3} M
  • Calculate pOH:
    pOH=−log(2.68×10−3)=2.57pOH = -log(2.68 \times 10^{-3}) = 2.57
  • Calculate pH:
    pH=14−2.57=11.43pH = 14 - 2.57 = 11.43