Easy Henderson-Hasselbalch problem

[Stacker]

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pH = pKa + log (/[a])

A patient requires 50 mg of active drug. Given in pill form, all of the drug is absorbed by the gut into the blood stream. If the active form is the protonated form, and the drug has a pKa of 6.8, how many milligrams must be given to ensure the proper dosage?

7 = 6.8 + log (/[a])
0.2 = log (/[a])
10^0.2 = /[a]
1.58/1 = /[a]
50/50 * 1.58/1 = 79/50 = /[a]
79 + 50 = 129 mg

Is 129 mg the correct answer?

Is it alright to give an extremely rough estimate of blood pH as 7.

 

[Pre-Med Village]

Blood pH is slightly alkaline ~ 7.4

This leads to a ratio of unprotonated to protonated form of about 4, so the answer should be about 250 mg.

 

[Stacker]

Blood pH is slightly alkaline ~ 7.4

This leads to a ratio of unprotonated to protonated form of about 4, so the answer should be about 250 mg.

Thanks!

 

[Morsetlis]

pH = pKa + log (/[a])

A patient requires 50 mg of active drug. Given in pill form, all of the drug is absorbed by the gut into the blood stream. If the active form is the protonated form, and the drug has a pKa of 6.8, how many milligrams must be given to ensure the proper dosage?

7 = 6.8 + log (/[a])
0.2 = log (/[a])
10^0.2 = /[a]
1.58/1 = /[a]
50/50 * 1.58/1 = 79/50 = /[a]
79 + 50 = 129 mg

Is 129 mg the correct answer?

Is it alright to give an extremely rough estimate of blood pH as 7.

It is not. 7.4.

 

[Sutton]

pH = pKa + log (/[a])

A patient requires 50 mg of active drug. Given in pill form, all of the drug is absorbed by the gut into the blood stream. If the active form is the protonated form, and the drug has a pKa of 6.8, how many milligrams must be given to ensure the proper dosage?

7.4 = 6.8 + log ( x / 50)
0.6 = log (x/50)
3.98 = x/50
199 = x
199mg + 50mg = 249 mg

 

[Morsetlis]

Only problem is, do they let you use a calculator with a log or inverse log function on the test?

 

[UrshumMurshum]

Only problem is, do they let you use a calculator with a log or inverse log function on the test?

Hopefully the answer choices aren't too close.

10^.6 is more than 10^.5 which is Sqrt(10). 3^2 is 9 so Sqrt(10) is probably somewhere around 3.2.

To account for the difference between 10^.5 and 10^.6 I'd probably adjust 3.2 to 3.5.

Anyone have a better way of approximating the exponent?

 

[Hexon]

pH = pKa + log (/[a])

A patient requires 50 mg of active drug. Given in pill form, all of the drug is absorbed by the gut into the blood stream. If the active form is the protonated form, and the drug has a pKa of 6.8, how many milligrams must be given to ensure the proper dosage?

7 = 6.8 + log (/[a])
0.2 = log (/[a])
10^0.2 = /[a]
1.58/1 = /[a]
50/50 * 1.58/1 = 79/50 = /[a]
79 + 50 = 129 mg

Is 129 mg the correct answer?

Is it alright to give an extremely rough estimate of blood pH as 7.

how did you get from

10^0.2 = /[a]

to

1.58/1 = /[a]

?

 

[Amphitrite]

10^0.2 = 1.58

 

[Hexon]

10^0.2 = 1.58

yeah, i know that, but seeing as how in the MCAT we're not allowed to use calculators, i'm curious to know what sort of mental shortcut the OP and others used to arrive at the answer?😛

 

[Erythropoietin]

yeah, i know that, but seeing as how in the MCAT we're not allowed to use calculators, i'm curious to know what sort of mental shortcut the OP and others used to arrive at the answer?😛

I don't think we are supposed to know complex calculations like that for the MCAT.

 

[Hexon]

I don't think we are supposed to know complex calculations like that for the MCAT.

phew, that's a relief lol i was just about to install a computer into my head😛

 

[Erythropoietin]

phew, that's a relief lol i was just about to install a computer into my head😛

You should be able to do log calculations though because I've seen a bunch in my practice exams...but they were never like 10^0.2. Just simple calculations of logs.

 

[bidiboom]

If you are not allowed to use calculator, you are not asked precise answers as well.. but they may demand logical thinking.. for instance there may be an answer that you can find with an approximate calculation. For instance;
10^0 = 1
10^1 = 10 .. so 10^0.2 must be between 1 and 10, and closer to 1 or 2.. but not 5 for instance, or 9.. so if there are 1.58 and 4.60 and 9.00 among the answers, you may be expected to see that the answer "should be" 1.58 ..

 

[Erythropoietin]

If you are not allowed to use calculator, you are not asked precise answers as well.. but they may demand logical thinking.. for instance there may be an answer that you can find with an approximate calculation. For instance;
10^0 = 1
10^1 = 10 .. so 10^0.2 must be between 1 and 10, and closer to 1 or 2.. but not 5 for instance, or 9.. so if there are 1.58 and 4.60 and 9.00 among the answers, you may be expected to see that the answer "should be" 1.58 ..

Yep that's what I meant. Well said.