midterm Q4

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Katherine Wu 1H
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midterm Q4

Postby Katherine Wu 1H » Wed Feb 19, 2020 5:32 pm

The drug amphetamine, C6H5CH2CH(CH3)NH2 (pKb=3.11), is usually marketed as the hydrogen bromide salt, C6H5CH2CH(CH3)NH3^+Br^-, because it is more stable in this solid form. When the drug is ingested as a salt (solid form), it enters the stomach, which contains digestive fluids at pH=1.7. Will this drug remain as C6H5CH2CH(CH3)NH3^+ or be converted to C6H5CH2CH(CH3)NH2? What is the ratio of the concentrations of the two species at this pH? Solve this problem be considering the salt in water.

I'm just confused as to why it remains C6H5CH2(CH3)NH3^+. Is it simply because the stomach is too acidic for it to become C6H5CH2CH(CH3)NH2?

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Re: midterm Q4

Postby 005391550 » Wed Feb 19, 2020 5:33 pm

The way i think of it is that the drug has a pKa of 10.89 and the "solution" it's in is pH 1.7 which is more acidic that the drug itself. because of that the drug will less likely give off protons and will act more like a base in the acidic surroundings.

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Re: midterm Q4

Postby BNgo_2L » Wed Feb 19, 2020 8:17 pm

You can determine which compound is present more by determining Ka for C6H5CH2CH(CH3)NH3+. Since the Ka = [products]/[reactants] and the Ka for the compound is extremely small, you can infer that there is a higher concentration of C6H5CH2CH(CH3)NH3+ (reactant) at equilibrium.

Brooke Yasuda 2J
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Re: midterm Q4

Postby Brooke Yasuda 2J » Wed Feb 19, 2020 8:51 pm

For this you have to look at the Ph of the solution and the PKa. Because the Ph is more acidic than the PKa, there will be no dissociation. Dissociation will only occur if the Ph is less acidic than the PKa because then the salt could dissociate and make the solution more acidic to reach the Pka value.

Bryan Chen 1H
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Re: midterm Q4

Postby Bryan Chen 1H » Wed Feb 19, 2020 10:57 pm

i think because the pH of the stomach is so acidic and there are already so many H+ floating around that the high pKa of the compound (and the natural dissociation of it) is too weak to "compete" with the already established acidity

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