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Calculations involving acidic, basic and buffer solutions

Calculations involving acidic, basic and buffer solutions. By Abdul- Wahhab Khedr Demonstrator of Pharmaceutics and Industrial Pharmacy. Calculation of hydrogen ion conc n [H + ] and hydroxyl ion conc n [OH - ] of weak acids and bases:. Weak acids:

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Calculations involving acidic, basic and buffer solutions

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  1. Calculations involving acidic, basic and buffer solutions By Abdul-WahhabKhedr Demonstrator of Pharmaceutics and Industrial Pharmacy

  2. Calculation of hydrogen ion concn [H+] and hydroxyl ion concn [OH-] of weak acids and bases: • Weak acids: The dissociation of weak acid can be expressed as follows: HB = H+ + B- Ka : is the acidity or dissociation constant of the weak acid. [H+]:is the hydrogen ion concentration resulting from dissociation of the weak acid. [B-]: is the concentration of the conjugate base of the weak acid resulting from dissociation and equals [H+] [HB]:is the concentration of the undissociated weak acid.

  3. Calculation of hydrogen ion concn [H+] and hydroxyl ion concn [OH-] of weak acids and bases: • Weak acids: Since the portion of the weak acid dissociated is very small compared to the undissociated portion, the concentration of undissociated weak acid [HB] equals the total acid concentration (C) Then, Hydrogen ion concn

  4. Calculation of hydrogen ion concn [H+] and hydroxyl ion concn [OH-] of weak acids and bases: • Weak bases: The dissociation of weak base can be expressed as follows: AOH = A+ + OH- Kb :is the basicity or dissociation constant of the weak base. [OH-]:is the hydroxyl ion concentration resulting from dissociation of the weak base. [A+]: is the concentration of the conjugate acid of the weak base resulting from dissociation and equals [OH-] [AOH]:is the concentration of the undissociated weak base.

  5. Calculation of hydrogen ion concn [H+] and hydroxyl ion concn [OH-] of weak acids and bases: • Weak bases: Since the portion of the weak base dissociated is very small compared to the undissociated portion, the concentration of undissociated weak base [AOH] equals the total base concentration (C) Then, Hydroxyl ion concn

  6. Calculation of hydrogen ion concn [H+] and hydroxyl ion concn [OH-] of strong acids and bases: Strong acids and bases are considered to be completely ionized when placed in water. For strong acid: [H+] ≈ C For strong base: [OH-] ≈ C

  7. Ionization constant and the hydrogen ion concn of water: The autoprotolytic reaction of water is represented as: H2O + H2O = H3O+ + OH- By application of law of mass action: K = [H3O+] . [OH-] / [H2O]2 Since the amount of water molecules dissociated is verysmall, the concentration of undissociated water molecules remain approximately constant. K × [H2O]2 = Kw = [H3O+] . [OH-] (Kw)is known as the dissociation constant, autoprotolysis constant or ion product of water. The value of the ion product of water Kw is1 × 10-14 Since, Kw = [H3O+]2 or [OH-]2 Then, [H3O+] = [OH-] = 1 × 10-7mole/L

  8. Conversion of hydrogen ion concn [H+] to pH • “pH” is the hydrogen ion exponent and is equal to the logarithm of the hydrogen ion concentration with a negative sign. pH = - log [H+] Similarly: pOH = - log [OH-] pKa= - log Ka pKb= - log Kb pKw= - log Kw And accordingly: pOH + pH = pKw pKa + pKb =pKw pKw= - log [1 × 10-14] = 14

  9. Conversion of hydrogen ion concn [H+] to pH For weak bases: [OH-] = (Kb .Cb)1/2 pOH = - log [OH-] pOH = - log (Kb.Cb)1/2 pOH = ½ (- log Kb + (-)log Cb) pOH = ½ (pKb + pCb) pH = pKw – pOH pH = pKw – ½ (pKb + pCb) For weak acids: [H+] = (Ka .Ca)1/2 pH = - log [H+] pH = - log (Ka .Ca)1/2 pH = ½ (- log Ka + (-)log Ca) pH = ½ (pKa + pCa)

  10. Examples • Calculate the [H+] and pH of 0.009 N hydrochloric acid solution. Solution: HCl is a strong acid (completely ionized) [H+] = 0.009 N pH = - log [H+] = - log 0.009 = 2.05 • Calculate the pH value of a solution of Sodium hydroxide whose [OH-] is 1.05 × 10-3 Solution: pOH = - log [OH-] = - log (1.05 × 10-3) = 2.98 pH = pKw – pOH = 14 – 2.98 = 11.02

  11. Examples • Calculate the hydrogen ion concentration of a solution of pH 5.3 Solution: pH = - log [H+] 5.3 = - log [H+] log [H+] = - 5.3 [H+] = antilog of (– 5.3) = 5.01 × 10-6 M • Calculate the pH and [H+] of 0.1 N acetic acid (pka = 4.76). Solution: pH = ½ (pKa + pCa) pH = ½ (pKa - log Ca) pH = ½ (4.76- log 0.1) = ½ (4.76 + 1) = ½ × 5.76 = 2.88 [H+] = antilog of (– 2.88) = 1.32 × 10-3 M

  12. Examples • Calculate the pH and [H+] of 0.13 N ammonia solution (pkb = 4.76). Solution: pH = pKw – ½ (pKb + pCb) pH = 14 – ½ (4.76 – log Cb) pH = 14 – ½ (4.76 – log 0.13) = 11.18 [H+] = antilog of (– 11.18) = 6.6 × 10-12 M

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