8.2: pH and pOH of Strong Acids and Bases

Strong Acids and Bases Strong Acids and Bases

• Strong acids and bases completely dissociate in water. This means that for every mole of strong acid/base added to water, one mole of $ H_3O^+ $ or $ OH^- $ ions are produced, respectively.
• Examples of strong acids: HCl, HBr, HI, $ HNO_3 $ , $ H_2SO_4 $ , $ HClO_4 $ .
• Examples of strong bases: Group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH), Group 2 hydroxides ( $ Ca(OH)_2 $ , $ Sr(OH)_2 $ , $ Ba(OH)_2 $ ).

pH Calculations

• pH is a measure of the acidity of a solution. It is defined as the negative logarithm (base 10) of the hydronium ion concentration ( $ [H_3O^+] $ ). $$ pH = -log[H_3O^+] $$ * For strong acids, the concentration of $ H_3O^+ $ is equal to the concentration of the strong acid, assuming complete dissociation.

  • If you have 0.1 M of HCl, then $ [H_3O^+] $ = 0.1 M.
  • $ pH = -log(0.1) = 1 $ .

pOH Calculations

• pOH is a measure of the basicity of a solution. It is defined as the negative logarithm (base 10) of the hydroxide ion concentration ( $ [OH^-] $ ). $$ pOH = -log[OH^-] $$ * For strong bases, the concentration of $ OH^- $ is equal to the concentration of the strong base, taking into account the stoichiometry.

  • If you have 0.05 M of NaOH, then $ [OH^-] $ = 0.05 M.
  • $ pOH = -log(0.05) = 1.30 $ .
  • If you have 0.05 M of $ Ba(OH)_2 $ , then $ [OH^-] = 2 \times 0.05M = 0.1 M $ .
  • $ pOH = -log(0.1) = 1 $ .

Relationship between pH and pOH

• In aqueous solutions at $ 25^\circ C $ , the sum of pH and pOH is always equal to 14. $$ pH + pOH = 14 $$ * This relationship is derived from the Ionic product of water Kw ( $ K_w $ ). $$ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} $$ $$ pH + pOH = -log(K_w) = -log(1.0 \times 10^{-14}) = 14 $$

Calculating $ [H_3O^+] $ and $ [OH^-] $ from pH and pOH

• To find the hydronium ion concentration from pH: $$ [H_3O^+] = 10^{-pH} $$ * To find the hydroxide ion concentration from pOH: $$ [OH^-] = 10^{-pOH} $$

Examples

1. Calculate the pH of a 0.020 M solution of $ HNO_3 $ .

   • $ HNO_3 $ is a strong acid, so $ [H_3O^+] = 0.020 M $ .
   • $ pH = -log(0.020) = 1.70 $ .

2. Calculate the pOH of a 0.0050 M solution of $ KOH $ .

   • $ KOH $ is a strong base, so $ [OH^-] = 0.0050 M $ .
   • $ pOH = -log(0.0050) = 2.30 $ .

3. Calculate the pH of a 0.0050 M solution of $ KOH $ .

   • $ KOH $ is a strong base, so $ [OH^-] = 0.0050 M $ .
   • $ pOH = -log(0.0050) = 2.30 $ .
   • $ pH = 14 - pOH = 14 - 2.30 = 11.70 $ .

4. Calculate the pH of a 0.01 M solution of $ Ca(OH)_2 $ .

   • $ Ca(OH)_2 $ is a strong base with two $ OH^- $ groups, so $ [OH^-]= 2 * 0.01 = 0.02 M $
   • $ pOH = -log(0.02)= 1.7 $
   • $ pH = 14 - pOH = 14 - 1.7 = 12.3 $

Strong Acids and Bases