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April 10th, 2018, 07:33 AM   #1
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Solubility question

A solution of sodium phenobarbital is to be prepared, given phenobarbital (MW = 232 g/mol, pKa = 7.4) and the molar solubility of phenobarbital is 1g/L.
(Note: MW sodium = 23 g/mol). Calculate the maximum concentration that can be achieved in solution (g/L) at pH 7.0.

My solution:
I used the weak acid solubility equation because phenobarbital is a weak acid.
pH = pKa + log ([S-So]/[So])
rearrange to get:
S = So (1+10^(pH-pKa))
In one of my lecture notes, it says that I have to use molar concentration (mol/L).

(1g/L)/(232g/mol) = 0.00431 mol/L
Substitute So = 0.00431 mol/L, pH = 7, pKa = 7.4 into equation to get:
0.00603moL/L. convert this to mol/L to get 1.40g/L

BUT FOR SOME REASON, the actual answer given is S=1.53g/L which is different to the answer i got. Would you be able to find out where I went wrong?
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April 10th, 2018, 07:36 AM   #2
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Quote:
molar solubility of phenobarbital is 1g/L.
How is the molar solubility measured in g per litre?
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April 10th, 2018, 07:43 AM   #3
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That is interesting. I'm not sure why. I double checked the question again and that is what it says.

If i assume that it is a typo, I still get 1.40g/L as the answer when i substitute it into the equation.
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April 10th, 2018, 07:47 AM   #4
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I will think about it, after I've checked you arithmetic post again later this evening.

You never know, you may be correct, if the author can make that sort of error.

I hope he/she never dispenses any medicine to me.
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April 10th, 2018, 07:53 AM   #5
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Haha, but I think i might be incorrect just because this is a test review question that all students did about a week ago. And since I was sick, I have to sit this test in 2 days time.

Usually, some students do notify the teacher when they think the answer is incorrect or is slightly different to their answer. But for this particular question, no has said anything, so I'm probably wrong.
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April 10th, 2018, 12:38 PM   #6
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You should have been alerted to the fact that they tell you the MW of sodium (=23)

This is because of this statement

Quote:
molar solubility of phenobarbital is 1g/L.
So this is the molar solubility of the phenobarbitone ion not the whole compound.

Do you understand that they do this because this is the controlling part of the compound for solubility - the sodium is many many times more soluble than this.

So the 1 gram refers to the phenobarb ion only (MW = 232 - 23).

So you have calculated that there are 1.4 (I made it 1.39 grams of phenobarb when the pH is 7.4.

So this must be multiplied by 232/209 to get the mass of sodium phenobarb in solution.


Does this help?


Good luck with your exam run up and the exam itself. I will look at your other question as soon as I can.
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