H-H Solutions - Exam 2

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What happens as you increase ionization?

Better for solubility, but worse for absorption

Acids - pH > pKa = ?

ionized

Bases - pH < pKa = ?

ionized

acid equation

pH - pKa = log (A-/HA) H+ + A- --> HA

Acids - pH < pKa = ?

unionized

What happens as you increase unionization?

- Better for absorption/membrane penetration, but worse for solubility - Not good for water solubility

(BASE EXAMPLE) What is the predominant form (ionized or unionized) of the molecule shown below in the stomach (pH=2) and in the duodenum (pH=4) and the ilium (pH=8). The drug has a pKa of 8.9 a) the stomach (pH = 2) b) the duodenum (pH = 4) c) the ilium (pH = 8) In which on elf these is the drug MOST soluble? In which of these will the drug be absorbed the most?

a) the stomach (pH = 2) 2 - 8.9 = log B/BH -6.9 = log B/BH 1/10^6.9 = B/BH 1/7,943,282 = B/BH ----> 1 B (base) has 7,943,282 BH (acid) = IONIZED __________________________________________________________________________ b) the duodenum (pH = 4) 4 - 8.9 = log B/BH -4.9 = log B/BH 1/10^4.9 = B/BH 1/79,432 = B/BH ----> 1 B (base) has 79,432 BH (acid) = IONIZED ___________________________________________________________________________ c) the ilium (pH = 8) 8 - 8.9 = log B/BH -0.9 = log B/BH 1/10^0.9 = B/BH 1/7.9 = B/BH ----> 1 B (base) has 7.9 BH (acid) = IONIZED - most soluble = duodenum ( high solubility and low absorption) - most absorbed = ilium (high absorption and low solubility)

base equation

pH - pKa = log (B/BH+) H+ + B <--> HB+

Bases - pH > pKa = ?

unionized

(ACID EXAMPLE) Aspirin (acetylsalicylic acid) has a pKA of 3.4. What is the ratio of A- to HA in: a) the blood (pH = 7.4) b) the stomach (pH = 1.4) c) the intestines (pH = 5.4)

a) the blood (pH = 7.4) 7.4 - 3.4 = log A-/HA 4 = log A-/HA 10^4 = A-/HA 10,000 = A-/HA ------> every 10,000 A- (acid) has 1 HA (base) = IONIZED _______________________________________________________________________________ b) the stomach (pH = 1.4) 1.4 - 3.4 = log A-/HA -2 = log A-/HA 10^-2 = A-/HA 1/10^2 = A-/HA 1/100 = A-/HA ------> every 1 A- (acid) has 100 HA (base) = UNIONIZED _______________________________________________________________________________ c) the intestines (pH = 5.4) 5.4 - 3.4 = log A-/HA 2 = log A-/HA 10^2 = A-/HA 100 = A-/HA ------> every 100 A- (acid) has 1 HA (base) = IONIZED


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