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An in-depth exploration of pH calculations, including the Brønsted-Lowry concept of acids and bases, conjugate acids and bases, pH and pOH values, and the calculation of pH for strong and weak acids and bases. It also covers titration and buffer systems.
Typology: Exams
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HCl(aq) + H 2 O(l) H 3 O+(aq) + Cl-(aq) Acid Base Conjugate acid
Conjugate base
H 2 O(l) + NH 3 (aq) NH 4 +(aq) + OH-(aq) Acid Base Conjugate acid
Conjugate base
Autoionization of water
Water is amphoteric as it can behave both as acid and base
2 H 2 O(l) H 3 O+(aq) + OH-(aq)
Kw = [H 3 O+][OH-] = [H+][OH-]
Ion-product constant for water:
In pure water at 25 ºC: [H+] = [OH-] = 1.0 × 10-7^ mol/L
Kw = (1.0 × 10-7^ mol/L)×( 1.0 × 10-7^ mol/L) = 1.0 × 10-14^ mol^2 /L^2
Constant!
Water contributes more protons than HCl in this case (10-7^ M), pH will be the same as in pure water, i.e. 7
Strong base
Inflection point:
If weak acid is just half-titrated, then pH = pK
[CH 3 COO−]. [H+] Kd = [CH 3 COOH]
Example: Calculate the pH of 0.01 mol/L acetic acid. Ka = 1.8 × 10-5.
pH = ½ × pK – ½ × log [AH]
pK = −log(1.8 × 10-5) = 4.
pH = ½ × 4.7447 − ½ × log0.01 = = 2.372 −(−1) = 3.
Example 2: Calculate the pH of 0.1 mol/L hypochlorous acid. Ka = 3.5 × 10-8.
pH = ½ × pK – ½ × log [AH]
pK = −log(3.5 × 10-8) = 7.
pH = ½ × 7.456 − ½ × log0.1 = = 3.728 −(−0.5) = 4.
Titration calculations
Example: An unknown sample of sulfuric acid H 2 SO 4 was titrated with the known KOH solution. It was found that 12 mL of the KOH c=0.1 mol/L was needed for just complete neutralisation of 10 mL H 2 SO 4 unknown sample.
What is concentration of sulfuric acid in the sample?
Equation: H 2 SO 4 + 2 KOH → K 2 SO 4 + 2 H 2 O
Calculation: H 2 SO 4 KOH c 1. v 1 = c 2. v 2
c 1 = c 2. v 2 / v 1 c 1 = 0.1. 12 / 10 = 0.
Including stoichiometry : c(H 2 SO 4 ) = 0.12/2 = 0.06 mol/L
Kd = [NH 3 ]
Example: Calculate the pH of 5 mol/L aqueous ammonia. Kb = 1.8 × 10-5.
pH = 14 – ½ × pKb + ½ × log [B]
pKb = −log(1.8 × 10-5) = 4.
pH = 14 − ½ × 4.7447 + ½ × log5 = = 14 − 2.37236 + 0.349 = 11.
Reaction of dissolved salts with water, e.g.:
A) Anion from a strong acid, cation from a weak base, e.g. NH 4 Cl: NH 4 +^ + H 2 O ↔ NH 3 + H 3 O+ Cl-^ + H 2 O … no reaction
B) Anion from a weak acid, cation from a strong base, e.g. NaHCO 3 : Na+^ + H 2 O … no reaction HCO 3 -^ + H 2 O ↔ H 2 CO 3 + OH-
…pH is acidic
…pH is alkaline
Buffers
Tris stands for: Tris(hydroxymethyl)aminomethane
pKa …negative log of dissociation constant of the weak acid [A–] …substance concentration of the salt/conjugate base [AH] …substance concentration of the weak acid
9:1 8:2 7:3 6:4 1:1 4:6 3:7 2:8 1:
5
6
7
8
9
10
Titration curve of sodium phosphate buffer
pH = pKa = 7.
How buffer works
E.g. phosphate buffer: NaH 2 PO 4 + Na 2 HPO 4
Addition of strong acid:
Na 2 HPO 4 + HCl → NaH 2 PO 4 + NaCl
Ionic: HPO 4 2-^ + H+^ → H 2 PO 4 -
Example 2: Calculate the pH of sodium phosphate buffer that originated from mixing 150 mL of 0.1 M NaH 2 PO 4 and 250 mL of 0.05 M Na 2 HPO 4. pKa = 7.21.
pH = pKa + log [A-]/[AH]
NaH 2 PO 4 : 0.1 mol/L × 0.15 = 0.015 mol
Na 2 HPO 4 : 0.05 mol/L × 0.25 = 0.0125 mol
pH = 7.21 + log 0.0125/0.015 = = 7.21 + (-0.07918) = = ~7.
Example 3: 500 mL of Tris free base 0.1 M was combined with 20 mL of 1 M HCl. What is predicted pH of the resulting Tris/HCl buffer? Tris pKb = 5.98.
pH = pKa + log [A-]/[AH] pKa = 14 – pKb
Tris free base: 0.05 mol HCl: 0.02 mol Tris + HCl → TrisHCl
pH = (14 – 5.98) + log (0.05-0.02)/0.02 = = 8.02 + 0.176 = = 8.196 = ~8.
