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Paracetamol contains a phenolic group (—OH) and the substance is therefore acidic. However, the pKa value is 9.7 , why is the pka value in basic?
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Mike Pollock
Paracetamol contains a phenolic group (—OH) and the substance is therefore acidic. However, the pKa value is 9.7 , why is the pka value in basic?
The other answers are all very complete and well explained.
Looking at things slightly differently, let’s ask “What is the pH of a 0.1M solution of paracetamol?”
The equilibrium will be something like HPa ←→ H+ + Pa-
Ka = [H+][Pa-]/[HPa] = 10^-pKa = 2X10^-10
Now, from the equilibrium, let [H+] = [Pa-] = x, and [HPa] = 0.1. Then,
2.0X10^-10 = x^2 / 0.1
x = [H+] = 4.4X10^-6
pH = 5.3
So, you see that the compound is, in fact, a weak acid.
The pKa merely tells you the pH at which the group is 50% ionised. For paracetamol the phenolic group is 50% ionised at pH 9.7 meaning paracetamol is a very weak acid. As a rule you have to be 2 log units greater than the pKa to get the pH at which you have 100% ionisation. For paracetamol you have to adjust the pH to 11.7 to get 100% ionisation of the phenolic group.
The pKa merely tells you the pH at which the group is 50% ionised. For paracetamol the phenolic group is 50% ionised at pH 9.7 meaning paracetamol is a very weak acid. As a rule you have to be 2 log units greater than the pKa to get the pH at which you have 100% ionisation. For paracetamol you have to adjust the pH to 11.7 to get 100% ionisation of the phenolic group.
The other answers are all very complete and well explained.
Looking at things slightly differently, let’s ask “What is the pH of a 0.1M solution of paracetamol?”
The equilibrium will be something like HPa ←→ H+ + Pa-
Ka = [H+][Pa-]/[HPa] = 10^-pKa = 2X10^-10
Now, from the equilibrium, let [H+] = [Pa-] = x, and [HPa] = 0.1. Then,
2.0X10^-10 = x^2 / 0.1
x = [H+] = 4.4X10^-6
pH = 5.3
So, you see that the compound is, in fact, a weak acid.
The other answers are all very complete and well explained.
Looking at things slightly differently, let’s ask “What is the pH of a 0.1M solution of paracetamol?”
The equilibrium will be something like HPa ←→ H+ + Pa-
Ka = [H+][Pa-]/[HPa] = 10^-pKa = 2X10^-10
Now, from the equilibrium, let [H+] = [Pa-] = x, and [HPa] = 0.1. Then,
2.0X10^-10 = x^2 / 0.1
x = [H+] = 4.4X10^-6
pH = 5.3
So, you see that the compound is, in fact, a weak acid.
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The pKa merely tells you the pH at which the group is 50% ionised. For paracetamol the phenolic group is 50% ionised at pH 9.7 meaning paracetamol is a very weak acid. As a rule you have to be 2 log units greater than the pKa to get the pH at which you have 100% ionisation. For paracetamol you have to adjust the pH to 11.7 to get 100% ionisation of the phenolic group.
The pKa merely tells you the pH at which the group is 50% ionised. For paracetamol the phenolic group is 50% ionised at pH 9.7 meaning paracetamol is a very weak acid. As a rule you have to be 2 log units greater than the pKa to get the pH at which you have 100% ionisation. For paracetamol you have to adjust the pH to 11.7 to get 100% ionisation of the phenolic group.
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