Polyprotic Acids and Mixtures of Acids Notes

Polyprotic Acids and Mixtures of Acids

Acid, Base, and Ampholyte Behavior

• Ampholytes (Amphoteric Compounds): Substances that can act as both acids and bases.

• Examples: "Acidic salts" of polybasic acids such as NaHS, NaH2PO4, Na2HPO4, and NaHCO3.

• Reactions:

• Acidic behavior: $H<em>3O^+ + S^{2-} \rightleftharpoons HS^- + H</em>2O \rightleftharpoons H_2S + OH^-$

• Acidic behavior: $H<em>3O^+ + HPO</em>4^{2-} \rightleftharpoons H<em>2PO</em>4^- + H<em>2O \rightleftharpoons H</em>3PO_4 + OH^-$

• Acidic behavior: $H<em>3O^+ + PO</em>4^{3-} \rightleftharpoons HPO<em>4^{2-} + H</em>2O \rightleftharpoons H<em>2PO</em>4^- + OH^-$

• Acidic behavior: $H<em>3O^+ + CO</em>3^{2-} \rightleftharpoons HCO<em>3^- + H</em>2O \rightleftharpoons H<em>2CO</em>3 + OH^-$

• The actual [H+] and pH of a solution is determined by considering all equilibria involving H+ ions.

pH Calculation of NaH2PO4 Solution

• Problem: Determine the pH of a 0.100 M NaH2PO4 solution.

• Given: For H3PO4:

  • $H<em>3PO</em>4(aq) \rightleftharpoons H^+(aq) + H<em>2PO</em>4^-(aq) K_{a1} = 7.5 \times 10^{-3}$

  • $H<em>2PO</em>4^-(aq) \rightleftharpoons H^+(aq) + HPO<em>4^{2-}(aq) K</em>{a2} = 6.2 \times 10^{-8}$

  • $HPO<em>4^{2-}(aq) \rightleftharpoons H^+(aq) + PO</em>4^{3-}(aq) K_{a3} = 1.0 \times 10^{-12}$

• Hydrolysis: Two possibilities:

  • Acidic hydrolysis (ionization):

    • $NaH<em>2PO</em>4(aq) \rightarrow Na^+(aq) + H<em>2PO</em>4^-(aq)$

    • $H<em>2PO</em>4^-(aq) \rightleftharpoons HPO_4^{2-}(aq) + H^+(aq)$

    • $K<em>{h,1} = K</em>{a,2} = \frac{[HPO<em>4^{2-}][H^+]}{[H</em>2PO_4^-]} = 6.2 \times 10^{-8}$

  • Base hydrolysis:

    • $H<em>2PO</em>4^-(aq) + H<em>2O \rightleftharpoons H</em>3PO_4(aq) + OH^-(aq)$

    • $K<em>{h,2} = \frac{[H</em>3PO<em>4][OH^-]}{[H</em>2PO<em>4^-]} = \frac{K</em>w}{K_{a,1}} = \frac{1.0 \times 10^{-14}}{7.5 \times 10^{-3}} = 1.3 \times 10^{-12}$

• Water Dissociation Equilibrium:

  • $H_2O \rightleftharpoons H^+(aq) + OH^-(aq)$

  • $K_w = [H^+][OH^-] = 10^{-14}$

• Since $K<em>{h,1} = 6.2 \times 10^{-8}$ and $K</em>{h,2} = 1.3 \times 10^{-12}$, neither hydrolysis reaction proceeds appreciably. Thus, the formal concentration of $[H<em>2PO</em>4^-]$ remains approximately 0.100 M.

• The first hydrolysis reaction proceeds better than the second, indicating the solution will be acidic.

Solving for [H+]

• Actual calculation:

  • $[H^+] = [H^+]<em>{from reactions (1 \&amp; 3)} - [H^+]</em>{used in reaction (2)}$

  • $[H^+] = \frac{K<em>{a,2}[H</em>2PO<em>4^-]}{[H^+]} + \frac{K</em>w}{[H^+]} - \frac{[H^+][H<em>2PO</em>4^-]}{K_{a,1}}$

  • Reorganization:

    • $[H^+] = [HPO<em>4^{2-}] + [OH^-] - [H</em>3PO_4]$

  • $[H^+] + \frac{[H^+]c}{K<em>{a,1}} = \frac{1}{[H^+]}(K</em>{a,2} \cdot c + K_w)$

  • $[H^+]^2 = \frac{K<em>{a,2} \cdot c + K</em>w}{1 + c/K_{a,1}}$

  • $[H^+]^2 = \frac{K<em>{a,1}(K</em>{a,2} \cdot c + K<em>w)}{K</em>{a,1} + c}$

pH Calculation

• $pH = \frac{1}{2}(pK<em>{a,1} + pK</em>{a,2}) + \frac{1}{2} \log \frac{K<em>{a,1} + c}{c + K</em>w/K_{a,2}}$

• Explicit Calculation:

  • $pH = \frac{1}{2}(2.12 + 7.21) + \frac{1}{2} \log \frac{7.5 \times 10^{-3} + 10^{-1}}{10^{-1} + 10^{-14}/6.8 \times 10^{-8}} = 4.66 + 0.01 = 4.67$

Simplified Relation for pH Calculation

• Conditions:

  • K_{a1} < c

  • Kw/K{a2} < c

• $pH = \frac{1}{2}(pK<em>{a,1} + pK</em>{a,2})$

• From $pH = \frac{1}{2}(pK<em>{a1} + pK</em>{a2}) = 4.66$

• In general, for ampholyte solutions, $pH = \frac{1}{2}(pK<em>a + pK</em>a')$

Titration of Weak Polyprotic Acid with Strong Base

• Titration Curve Properties:

  • In each buffer zone: $pH = pK_a + \log \frac{[corresponding Base]}{[corresponding Acid]}$

  • Halfway the buffer zone, where $[corresponding Base] = [corresponding Acid]$, thus $pH = pK_a$

  • pH-jump at each stoichiometric point (if successive acid constants differ sufficiently).

  • At the stoichiometric point between two buffer zones, an ampholyte is present, and the pH is approximated by $pH = \frac{1}{2}(pK<em>a + pK</em>a')$

Application: Titration Curve of H3PO4 with NaOH

• Titration of 40 mL 0.01 M H3PO4 with 0.05 M NaOH.

• Given: $pK<em>{a1} = 2.12$, $pK</em>{a2} = 7.20$, $pK_{a3} = 12.00$

• Steps:

  1. 0 to 1 stoichiometric amount of NaOH:

     • Reaction: $H<em>3PO</em>4(aq) + OH^-(aq) \rightarrow H<em>2PO</em>4^-(aq) + H_2O$

     • Buffer mixture: H3PO4 / H2PO4-

     • At 0.5 stoichiometric amounts: $[H<em>3PO</em>4] = [H<em>2PO</em>4^-]$ and $pH \approx pK_{a,1} = 2.12$

  2. 1 to 2 stoichiometric amount of NaOH:

     • Reaction: $H<em>2PO</em>4^-(aq) + OH^-(aq) \rightarrow HPO<em>4^{2-}(aq) + H</em>2O$

     • Buffer mixture: H2PO4- / HPO42-

     • At 1.5 stoichiometric amounts: $[H<em>2PO</em>4^-] = [HPO<em>4^{2-}]$ and $pH \approx pK</em>{a,2} = 7.20$

  3. 2 to 3 stoichiometric amount of NaOH:

     • Reaction: $HPO<em>4^{2-}(aq) + OH^-(aq) \rightarrow PO</em>4^{3-}(aq) + H_2O$

     • Buffer mixture: HPO42- / PO43-

     • At 2.5 stoichiometric amounts: $[HPO<em>4^{2-}] = [PO</em>4^{3-}]$ and $pH \approx pK_{a,3} = 12.00$

  4. At the 1st stoichiometric point:

     • H2PO4- is present (ampholyte).

     • $pH = \frac{1}{2}(pK<em>{a1} + pK</em>{a2})$

  5. At the 2nd stoichiometric point:

     • HPO42- is present (ampholyte).

     • $pH = \frac{1}{2}(pK<em>{a2} + pK</em>{a3})$

pH of Mixtures of Acids and Bases

1. Solutions of Two Strong Acids (or Bases)

• Since both acids (or bases) are fully dissociated:

  • Mixture of 2 strong acids: $[H^+] = [H^+]<em>{acid,1} + [H^+]</em>{acid,2}$

  • Mixture of 2 strong bases: $[OH^-] = [OH^-]<em>{base,1} + [OH^-]</em>{base,2}$

• Example: 0.010 mol HCl and 0.020 mol HNO3 in 500 mL water.

  • $[H^+]_{HCl} = \frac{0.01 mol}{0.5 L} = 0.020 M$

  • $[H^+]<em>{HNO</em>3} = \frac{0.02 mol}{0.5 L} = 0.040 M$

  • $[H^+] = 0.020 M + 0.040 M = 0.060 M$

  • $pH = -\log[H^+] = 1.22$

  • Actual concentrations: $[Cl^-] = 0.020 M$, $[NO_3^-] = 0.040 M$, $[H^+] = 0.060 M$

2. Solutions of Strong Acid and Weak Acid (or Strong Base and Weak Base)

• Since the strong acid is dissociated completely and the weak acid almost not: [H^+]{strong acid} >> [H^+]{weak acid}

• Approximation: $[H^+] = [H^+]_{strong acid}$

• Example: 0.010 M HCl and 0.010 M HOAc (pKa = 4.75).

  • HCl = strong acid, HOAc = weak acid

  • $[H^+] = [H^+]_{HCl} = 0.01 M$

  • $pH = -\log[H^+] = 2.00$

  • Formal concentrations: 0.010 M HCl, 0.010 M HOAc

  • Actual concentrations: $[H^+] = 0.010 M$, $[Cl^-] = 0.010 M$, $[HOAc] = 0.010 M$

3. Solutions of Mixtures of Acids and Bases

• pH determination in two steps:

  1. Mixing and ending reactions (stoichiometric calculations using moles).

  2. Equilibrium calculations (hydrolysis equilibrium, [H+] calculation using concentrations).

Step 1

• React actual contributed acids and bases stoichiometrically with each other (ending reactions).

Step 2

• Write the hydrolysis equilibrium in water for the resulting end product(s) and calculate the [H+] with the associated equilibrium constant.
  Example: 10 mL 0.20 M NH3, 20 mL 0.20 M NH4NO3, 30 mL 0.10 M HCl, 40 mL 0.05 M NaOH are mixed. Given $pK<em>a(NH</em>4^+) = 9.24$

• Identify Brönsted-acids and bases and their amounts:

  • NH3 (weak base): 10 mL x 0.20 mol/L = 2.0 mmol

  • NH4+ (weak acid): 20 mL x 0.20 mol/L = 4.0 mmol

  • H+ (strong acid): 30 mL x 0.10 mol/L = 3.0 mmol

  • OH- (strong base): 40 mL x 0.05 mol/L = 2.0 mmol

• Ending reactions (strong with strong > strong with weak > weak with weak):

  • a) $H^+(aq) + OH^-(aq) \rightarrow H_2O$

    • Initial: 3.0 mmol 2.0 mmol

    • Reaction: -2.0 mmol -2.0 mmol

    • Remains: 1.0 mmol -

  • b) $NH<em>3(aq) + H^+(aq) \rightarrow NH</em>4^+(aq)$

    • I: 2.0 mmol 1.0 mmol 4.0 mmol

    • C: -1.0 mmol -1.0 mmol +1.0 mmol

    • E: 1.0 mmol - 5.0 mmol

• In the 100 mL solution, there is:

  • 1.0 mmol NH3 $[NH_3] = 0.010 M$

  • 5.0 mmol NH4+ $[NH_4^+] = 0.050 M$

• This generates a buffer.

Step 2: Equilibrium Calculation

• Possible equilibrium: $NH<em>4^+(aq) \rightleftharpoons NH</em>3(aq) + H^+(aq)$

• Buffer: 0.050 M 0.010 M ???

  • $K<em>a = \frac{[NH</em>3][H^+]}{[NH_4^+]}$

  • $pH = pK<em>a + \log \frac{[NH</em>3]}{[NH_4^+]} = 9.24 + \log \frac{0.01 M}{0.05 M} = 8.54$

• Conclusion: 1.0 mmol NH3, 5.0 mmol NH4+, 4.0 mmol NO3-, 3.0 mmol Cl-, 2 mmol Na+ in 100 mL solution.

Example 2

Solution which are mixed together: 10 mL 0,30 M NH3 20 mL 0,30 M NH4Cl 30 mL 0,10 M HCl 40 mL 0,10 M HOAc
What is the final pH of the mixture?
step 1:

• Which Brönsted-acids and bases, and how much mols are present? NH3 : weak base: 10 mL x 0,30 mol.L-1 = 3,0 mmol NH4 +: weak acid: 20 mL x 0,30 mol.L-1 = 6,0 mmol H+: strong acid: 30 mL x 0,10 mol.L-1 = 3,0 mmol HOAc: weak acid: 40 mL x 0,10 mol.L-1 = 4,0 mmol

  • Ending reactions (sequence: strong with strong > strong with weak > weak with weak):

a) NH3(aq) + H+(aq) ® NH4 +(aq)
I 3,0 3,0 6,0 (mmol)
C -3,0 -3,0 +3,0 (mmol)
E - - 9,0 (mmol)
All base is used rest: unreacted acid HOAc and NH4 +
In this 100 mL solution is actually present :
9,0 mmol NH4 + (weak acid) (Ka=10-9,24)
4,0 mmol HOAc (weak acid) (Ka=10-4,74) ®The pH of the solution is only determined by the strongest acid!
Stronger acid!
[HOAc] = 4,0 mmol 100 mL = 0,04 M

step 2 : equilibrium in water :
HOAc(aq) OAc-(aq) + H+(aq)
formal 0,04 M (in mol.L-1)
equilibrium (0,04 - x) M x M x M
K a = 10-4,74 = [H+] [OAc -] [HOAc] = x 2 0,04 - x = 10-4,74 or x 2 + 10 -4,74 x - 0,04.10 -4,74 = 0
So x = [H+] = 8,44 x 10-4 and pH = 3,07
Formal : 3,0 mmol NH3 6,0 mmol NH4 + (and Cl-) 3,0 mmol H + (and Cl-) 4,0 mmol HOAc in 100 mL solution
Actually present: 9,0 mmol NH4 + (and NO3 -) 9,0 mmol Cl- 4,0 mmol HOAc in 100 mL solution
Conclusion