Help!! Chemistry?

Question

I'm doing a pharmacology experiment and am kind of confused using log equations. The question is determine what the percentage of drug X in the ionised form would be at pH 5 when the pKa of drug X = 5.8. I'm confused with how to get the percentage from the Henderson-Hasselbach equation.

Answer 1

The H-H eqn states that the following equality holds for an acidic system:
pH = pKa + log([A-]/[HA])
where A- is the conjugate base of the acid HA. If you look at the equation closely, the term [A-]/[HA] is the ratio of the ionized form of the drug to the unionized form. However, if we assume X to be a weak acid, we can assume that the unionized form of the drug is approximately equal to the original amount, or its formal concentration, of the drug in the solution. Therefore, the term [A-]/[HA] can be used to give a fairly accurate value of the ratio of the drug that ionized.

Solving for [A-]/[HA]:
[A-]/[HA] = 10^(pH-pKa)
Substituting the given values,
[A-]/[HA] = 0.16
or around 16%. This kind of justifies our assumption that X is a weak acid. You can also do the calculation done by the guy above me but you'd find the answer is 15%, not 1.5%.

Answer 2

let drug molecule be HA; when it ionises

HA---> H+ + A-

==> Ka =
[H+] [A-]
------------ (division sign)
[HA]

[H+] = [A-] ( [ ] in the notation i'm using for concentration.)


HA---> H+ + A-

==> Ka =
[H+]^2
------------
[HA]

pKa= - log Ka
pH = -log[H+]

From these eqn you can find Ka, concentration of H+ which is equal to concentration of drug ionised & also concentration of HA(drug in the solution). Then calculate % from conc. of H+ & that of HA.
my calculation says [H+] = 10^-5 moldm-3
& ka= 10^-5.8 moldm-3
[HA]=(((10^-5)^2)/(10^-5.8))=6.3x10^-5 moldm-3
==> %= ((10^-5)/(6.3x10^-5))x100= 16%

Answer 3

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Answer 4

No one is happy with log equations, so dont use them. Assuming this is a monobasic acid HX, you can write the equilibrium of the acid with it's ionized form as:
[X-][H+]/[HX] = Ka.
Ka = 10^-5.8 or 10^-6 * 10^0.2 or appx 1.5x10^-6
[H+] is 1x10^-5. So you can find [X-]/[HX]= 0.015, or in percentages, 1.5%

Answer 5

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