Dosage Form Design 1 2023 Question 3 (cont.) 2) Test 1 Why not strong ace? Calculate the pH of an aqueous solution of 1% ephedrine HCI (MW = 202 g/mol, pKa = 9.6) ephedrine HCI => conjugate weak acid of the weak base ephedrine concentration 1 % (g/100 mL) MW 202 g/mol concentration 0.0495 M =%*10/MW =1*10/202 ρκα 9.6 pH=0.5pKa-0.5*log([C]) PH 5.5 DDD.
Dosage Form Design 1 2023 Question 3 (cont.) 2) Test 1 Why not strong ace? Calculate the pH of an aqueous solution of 1% ephedrine HCI (MW = 202 g/mol, pKa = 9.6) ephedrine HCI => conjugate weak acid of the weak base ephedrine concentration 1 % (g/100 mL) MW 202 g/mol concentration 0.0495 M =%*10/MW =1*10/202 ρκα 9.6 pH=0.5pKa-0.5*log([C]) PH 5.5 DDD.
Chapter14: Formula Method
Section: Chapter Questions
Problem 5SST
Related questions
Question
I cannot understand why the ephedrine HCl is considered as a conjugated weak acid of the weak base ephedrine. I think I am missing out some concepts. I know the ibuprofen is a weak acid and ibuprofen sodium is a weak base I just memorised it cuz I cannot understand why. I only know sodium salt of wa acts a s wb. Sodium salt of wb acts as a weak acid
![Dosage Form Design 1
2023
Question 3 (cont.)
2)
Test 1
Why not strong
ace?
Calculate the pH of an aqueous solution of 1% ephedrine HCI (MW = 202 g/mol, pKa = 9.6)
ephedrine HCI
=> conjugate weak acid of the weak base ephedrine
concentration
1 % (g/100 mL)
MW
202
g/mol
concentration
0.0495
M
=%*10/MW
=1*10/202
ρκα
9.6
pH=0.5pKa-0.5*log([C])
PH
5.5
DDD.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcccd0f27-89d9-42a1-bb7f-6b052790125e%2F5725ea87-4da7-46ac-9e11-21ac3f25137a%2Fuf0tdng_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Dosage Form Design 1
2023
Question 3 (cont.)
2)
Test 1
Why not strong
ace?
Calculate the pH of an aqueous solution of 1% ephedrine HCI (MW = 202 g/mol, pKa = 9.6)
ephedrine HCI
=> conjugate weak acid of the weak base ephedrine
concentration
1 % (g/100 mL)
MW
202
g/mol
concentration
0.0495
M
=%*10/MW
=1*10/202
ρκα
9.6
pH=0.5pKa-0.5*log([C])
PH
5.5
DDD.
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