Acids, Bases, & Electrolytes

Acids, Bases, & Electrolytes

Application

  1. Increase or influence solubility
    • Henderson-Hasselbach
    • pH = pKa + log base/acid
  2. Maintain a non-irritating pH
  3. Drug Stability
  4. Drug Activity
  5. Drug Absorption

  1. Increase in Solubility
    - many drugs are either acid or bases and their solubility depend on pH of the solution (see solubility graph).

  2. Maintain a Non-irritating pH
    • Lacrymal fluid has a pH = 7.4 and eye can tolerate pHıs 5-9 because:
      • small volume of solution is being introduced
      • increase production of tears
      • buffer of lacrymal fluid
    • Parenteral
      • isotonic condition
      • depend on volume, acidity and basicity
      • end time for delivery
    • Buffer capacity of ophthalmics and parenterals should be low so body can readily adjust to them.
  3. Drug Stability
    • Catalytic effect of acids and bases on hydrolysis or oxidation (see graph of pH vs stability)
    • Atropine
      • maximum stability at pH 3.8
      • phosphate buffer at pH 6.8 often recommended for irritation, absorption and stability reason
    • Drug Activity
      • Some drugs are active in non-ionized form
        For Example:

      • Some antibacterial are effective only in cationic form
        For example: Acridines - quats
    • Drug Absorption

      - Degree of ionization and lipd solubility
      • Only a non-ionized species of a drug can cross cellular membrane
      • Weak acid absorbed better from an acid media and weak bases from a basic media
      • Rate of absorption of drugs often related to its pKa
        - Amines and alkaloids more effective at higher pH




      - Treatment of amphetamines, barbiturates and narcotics overdoses
      • pH of urine was adjusted to increase the excretion of the ionized species.

      - Objective is to balance the effects of pH and buffers

Theory of Acid & Base


  1. Arrhenius:
    • Acid liberates H+ in water
    • Base liberates OH- in water
  2. Bronsted-Lowry:
    • Acid is a proton donor
    • Base is a proton acceptor


  3. Lewis Acid and Base:
    • Acid is electron pair acceptor
    • Base is electron pair donor




A. Strong Acid

DISSOCIATE COMPLETELY



B. Strong Base

    DISSOCIATE COMPLETELY



    Examples:

    Calculate pH of solution with the following hydrogen ion cencetration:

  1. [H+] = 4.57 x 10-7

    pH = - log [H+] = - log 4.57 x 10-7

    pH = - (log 4.57 + log 10-7)

    pH = - log 4.57 - log 10-7

    pH = - 0.66 + 7 = 6.34

  2. [H+] = 8.43 x 10-2

    pH = - log [H+] = - log 8.43 x 10-2

    pH= - log 8.43 - log 10-2

    pH = - 0.93 + 2 = 1.07

  3. Calculate pOH and pH of 5.0 x 10-2 M solution of NaOH. NaOH ------------> Na+ + Cl-

    5.0 x 10-2M 5.0 x 10-2 M 5.0 x 10-2M

    [OH-] = 5.0 x 10-2 M

    pOH = - log 5 - log 10-2

    pOH = 0.70 - 2.0 = 1.3

    pH = 14.0 - 1.30 = 12.7

  4. Calculate the pH of solution preapred by mixing 2.0 mL of strong acid (HCl) of pH 3.0 and 3.0 mL of strong base (NaOH) of pH 10.0. Strong Acid: HCl ----------> H+ + Cl-
    ? ?

    pH = 3.0

    - log [H+] = 3.0

    [H+] = 1 x 10-3 M = 1 x 10- 3 mole/1

    [H+] = 1 x 10-3 mmole/mL

    [H+] in 2 mL of solution contains =
    2 mL x 10-3 mmole/mL = 2 x 10-3 mmole of H+

    Strong Base:
    NaOH ----------> Na+ + OH-

    pH = 10

    pOH = 14 - 10 = 4

    [OH-] = 1 x 10-4M = 1 x 10- 4 mmole/mL

    3 mL of NaOH (base) = 3 mL x 1 x 10-4 mmole/mL = 3 x 10-4 mmole of OH-
    H+ + OH- -------- > H2O

    2 x 10-3 mmole 3 x 10-4 mmole 3 x 10-4 mmole

    Excess Acid (HCl) = 2 x 10-3 mmole - 3 x 10-4 mmole

    [H+] = 0.0017 mmole H+ in 5 mL solution

    [H+]= 0.0017 mmole/5 mL

    [H+] = 3.4 x 10-4 mmole/mL

    [H+] = 3.4 x 10-4 M

    pH = - log 3.4 x 10-4

    pH = 3.47

  5. Other way:

    HCl before mixing with NaOH:
    M1 = [HCl] = 1 x 10-3 M
    Volume = VHCl = V1 = 2 mL
    HCl after mixing with NaOH:
    V2 = 5 mL
    M2 = [HCl] = ?
    V1 x M1 = V2 x M2
    2 mL x 1 x 10-3 M = 5 mL x M2
    M2 = (2 mL x 1 x 10-3 M)/5 mL = 4 x 10-4 M = [HCl]

    NaOH before mixing with HCl:
    M1 = [NaOH] = 1 x 10-4 M
    Volume = VNaOH = V1 = 3 mL

    NaOH after mixing with HCl:
    V2 = 5 mL
    M2 = [NaOH] = ?
    V1 x M1 = V2 x M2
    3 mL x 1 x 10-4 M = 5 mL x M2
    M2 = (3 mL x 1 x 10-4 M) / 5 mL= 6 x 10-5 M = [NaOH]

    HCl + NaOH --------> NaCl + H2O
    4 x 10-4M 6 x 10-5 M 6 x 10-5M 6 x 10-5 M

    Left over HCl = 4 x 10-4 M - 6 x 10-5 M

    [H+] = 3.4 x 10-4 M

    pH = - log 3.4 x 10-4 M = 3.47



C. Weak Acid






D. Weak Base:





E. Salt of Weak Acid:




F. Salt of Weak Base

BH+ =======> B + H+

[H+] = KaCsalt

pH = 1/2 (pKa - log Csalt)

Example:
Calculate pH of 0.25 M solution of NH4Cl.
Kb = 1.75 x 10-5 NH4Cl ====== NH4+ + Cl-

NH4+ ======= NH3 + H +

Ka = Kw/Kb = 10 -14 /1.75 x 10 -5

Ka = 5.71 x 10 -10

pH = 1/2 (-log Ka - log Csalt)

pH = 1/2 (-log 5.71 x 10 -10 - log 0.25)

pH = 1/2 (9.24 + 0.60) = 4.92



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