1. Calculate the pH of a 0.01 M Solution of salicylic acid which has a K a =1.06
ID: 897260 • Letter: 1
Question
1. Calculate the pH of a 0.01 M Solution of salicylic acid which has a Ka=1.06×10-3.
2. Calculate the pH of a 1g/100mL solution of ephedrine sulfate. The molecular weight of the salt is 428.5 and Kb for ephedrine base is 2.3×10-5.
3. The ionization constant Kb for morphine base is 7.4×10-7. What is the hydroxyl ion concentration of a 0.0005 M aqueous solution of morphine?
4. What is the pH of a 0.0033 M solution of cocaine base, which has a basicity constant of 2.6×10-6?
5. Calculate the pH of a 0.165 M solution of sodium sulfathiozole. The acidicity constant for sulfathizone is 7.6×10-8.
I need help with these 5 questions!!!
Explanation / Answer
1.Calculate the pH of a 0.01 M Solution of salicylic acid which has a Ka=1.06×10-3.
pKa = -log Ka = -log (1.06×10-3.) = 2.97
pH = 1/2 (pKa - log C)
= 1/2 (2.97 - log 0.01)
= 2.48
pH = 2.48
2)
2. Calculate the pH of a 1g/100mL solution of ephedrine sulfate. The molecular weight of the salt is 428.5 and Kb for ephedrine base is 2.3×10-5
molarity (C) = ( mass /molar mass ) x 1000 / 100
= (1/ 428.5) x 1000 / 100
= 0.0233 M
Kb = 2.3×10-5
pKb = -log Kb = -log ( 2.3×10-5) = 4.64
ephedrine sulfate it is the salt of weak base and stron acid . so the pH should be less than 7
pH = 7 - 1/2 [pKb + log C]
pH = 7 - 1/2 [4.64 + log 0.0233]
pH = 5.5
3. ) The ionization constant Kb for morphine base is 7.4×10-7. What is the hydroxyl ion concentration of a 0.0005 M aqueous solution of morphine
Kb = 7.4 x 10^-7
C = 0.0005 M
[OH-] = sqrt (Kb x C)
= sqrt (7.4 x 10^-7 x 0.0005)
= 1.92 x 10^-5 M
4. What is the pH of a 0.0033 M solution of cocaine base, which has a basicity constant of 2.6×10-6?
pkb = -log (2.6×10-6) = 5. 58
pOH = 1/2 [pKb + log C] = 1/2 [5.58 + log 0.0033] = 1.55
pH + pOH = 14
pH = 12.45
5. Calculate the pH of a 0.165 M solution of sodium sulfathiozole. The acidicity constant for sulfathizone is 7.6×10-8.
Ka = 7.6×10-8
pKa = 7.12
pH = 7 + 1/2 [pKa + logC]
pH = 10.17
pH = 7+ 1/2 [7.12 + log 0.165]
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