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: H3O^+ + S^{2-} \rightleftharpoons HS^- + H2O \rightleftharpoons H_2S + OH^-
Acidic behavior: H3O^+ + HPO4^{2-} \rightleftharpoons H2PO4^- + H2O \rightleftharpoons H3PO_4 + OH^-
Acidic behavior: H3O^+ + PO4^{3-} \rightleftharpoons HPO4^{2-} + H2O \rightleftharpoons H2PO4^- + OH^-
Acidic behavior: H3O^+ + CO3^{2-} \rightleftharpoons HCO3^- + H2O \rightleftharpoons H2CO3 + 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:
H3PO4(aq) \rightleftharpoons H^+(aq) + H2PO4^-(aq) K_{a1} = 7.5 \times 10^{-3}
H2PO4^-(aq) \rightleftharpoons H^+(aq) + HPO4^{2-}(aq) K{a2} = 6.2 \times 10^{-8}
HPO4^{2-}(aq) \rightleftharpoons H^+(aq) + PO4^{3-}(aq) K_{a3} = 1.0 \times 10^{-12}
Hydrolysis: Two possibilities:
Acidic hydrolysis (ionization):
NaH2PO4(aq) \rightarrow Na^+(aq) + H2PO4^-(aq)
H2PO4^-(aq) \rightleftharpoons HPO_4^{2-}(aq) + H^+(aq)
K{h,1} = K{a,2} = \frac{[HPO4^{2-}][H^+]}{[H2PO_4^-]} = 6.2 \times 10^{-8}
Base hydrolysis:
H2PO4^-(aq) + H2O \rightleftharpoons H3PO_4(aq) + OH^-(aq)
K{h,2} = \frac{[H3PO4][OH^-]}{[H2PO4^-]} = \frac{Kw}{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{h,1} = 6.2 \times 10^{-8} and K{h,2} = 1.3 \times 10^{-12}, neither hydrolysis reaction proceeds appreciably. Thus, the formal concentration of [H2PO4^-] 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^+]{from reactions (1 \& 3)} - [H^+]{used in reaction (2)}
[H^+] = \frac{K{a,2}[H2PO4^-]}{[H^+]} + \frac{Kw}{[H^+]} - \frac{[H^+][H2PO4^-]}{K_{a,1}}
Reorganization:
[H^+] = [HPO4^{2-}] + [OH^-] - [H3PO_4]
[H^+] + \frac{[H^+]c}{K{a,1}} = \frac{1}{[H^+]}(K{a,2} \cdot c + K_w)
[H^+]^2 = \frac{K{a,2} \cdot c + Kw}{1 + c/K_{a,1}}
[H^+]^2 = \frac{K{a,1}(K{a,2} \cdot c + Kw)}{K{a,1} + c}
pH Calculation
pH = \frac{1}{2}(pK{a,1} + pK{a,2}) + \frac{1}{2} \log \frac{K{a,1} + c}{c + Kw/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{a,1} + pK{a,2})
From pH = \frac{1}{2}(pK{a1} + pK{a2}) = 4.66
In general, for ampholyte solutions, pH = \frac{1}{2}(pKa + pKa')
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}(pKa + pKa')
Application: Titration Curve of H3PO4 with NaOH
Titration of 40 mL 0.01 M H3PO4 with 0.05 M NaOH.
Given: pK{a1} = 2.12, pK{a2} = 7.20, pK_{a3} = 12.00
Steps:
0 to 1 stoichiometric amount of NaOH:
Reaction: H3PO4(aq) + OH^-(aq) \rightarrow H2PO4^-(aq) + H_2O
Buffer mixture: H3PO4 / H2PO4-
At 0.5 stoichiometric amounts: [H3PO4] = [H2PO4^-] and pH \approx pK_{a,1} = 2.12
1 to 2 stoichiometric amount of NaOH:
Reaction: H2PO4^-(aq) + OH^-(aq) \rightarrow HPO4^{2-}(aq) + H2O
Buffer mixture: H2PO4- / HPO42-
At 1.5 stoichiometric amounts: [H2PO4^-] = [HPO4^{2-}] and pH \approx pK{a,2} = 7.20
2 to 3 stoichiometric amount of NaOH:
Reaction: HPO4^{2-}(aq) + OH^-(aq) \rightarrow PO4^{3-}(aq) + H_2O
Buffer mixture: HPO42- / PO43-
At 2.5 stoichiometric amounts: [HPO4^{2-}] = [PO4^{3-}] and pH \approx pK_{a,3} = 12.00
At the 1st stoichiometric point:
H2PO4- is present (ampholyte).
pH = \frac{1}{2}(pK{a1} + pK{a2})
At the 2nd stoichiometric point:
HPO42- is present (ampholyte).
pH = \frac{1}{2}(pK{a2} + pK{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^+]{acid,1} + [H^+]{acid,2}
Mixture of 2 strong bases: [OH^-] = [OH^-]{base,1} + [OH^-]{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^+]{HNO3} = \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:
Mixing and ending reactions (stoichiometric calculations using moles).
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 pKa(NH4^+) = 9.24Identify 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) NH3(aq) + H^+(aq) \rightarrow NH4^+(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: NH4^+(aq) \rightleftharpoons NH3(aq) + H^+(aq)
Buffer: 0.050 M 0.010 M ???
Ka = \frac{[NH3][H^+]}{[NH_4^+]}
pH = pKa + \log \frac{[NH3]}{[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