I am unsure if I am doing this right.
The question is: Aspirin (acetylsalicylic acid) can cause stomach bleeding. Only the acid form can cross the stomach lining. Determine the ratio of base to its conjugate acid when the stomach pH=2.00. Ka aspirin = 2.75x10^-5.
I used pH = pKa - log([HA]/[A])
pKa = -logKa
= -log(2.7x10^-5)
=4.56
2.0 = 4.56 + log(2.00/2.7x10^-5)
2.0 = 9.43
divide both by 2.0 and got 4.715
Is this right?
I didn't check the arithmetic but you started with the correct equation although I prefer pH = pKa + log (A^-/HA)
It appears to me that you didn't substitute correctly for HA/A. You apparently substitute pH and Ka for that fraction. What you want to do is to solve for the fraction; i.e.,
pH = pKa + log [(A^-)/(HA)]
2.00 = 4.56 + log base/acid
2.00-4.56 = log base/acid
-2.56 = log (base/acid)
Solve for base/acid which is (A^-)/(HA)
To determine the ratio of base to its conjugate acid when the stomach pH is 2.00, you have correctly used the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
First, you correctly calculated the pKa value for aspirin:
pKa = -log(Ka)
pKa = -log(2.75x10^-5)
pKa = 4.56
Next, you plugged in the given pH and pKa values into the Henderson-Hasselbalch equation and solved for the ratio [A-]/[HA]:
2.00 = 4.56 + log(2.00/[HA])
-2.56 = log(2.00/[HA])
To solve for [HA], you need to exponentiate both sides of the equation:
10^(-2.56) = [2.00/[HA]]
0.00355 = 2.00/[HA]
Now, solving for [HA]:
[HA] = 2.00 / 0.00355
[HA] ≈ 563.38
Therefore, the ratio of [A-] to [HA] (base to conjugate acid) when the stomach pH is 2.00 is approximately 563.38:1.
To determine the ratio of base to its conjugate acid, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
In this case, the base is the conjugate base of aspirin (acetylsalicylic acid), and the acid is the aspirin itself.
First, you correctly calculated the pKa value for aspirin:
pKa = -log(Ka)
= -log(2.75x10^-5)
≈ 4.56
Next, substitute the values into the Henderson-Hasselbalch equation:
2.00 = 4.56 + log([base]/[acid])
Rearrange the equation:
log([base]/[acid]) = 2.00 - 4.56
log([base]/[acid]) ≈ -2.56
Take the antilog of both sides:
[base]/[acid] ≈ 10^(-2.56)
Using the inverse logarithm function on a calculator, you can find that 10^(-2.56) is approximately 3.60 x 10^(-3).
Therefore, the ratio of the base to the conjugate acid when the stomach pH is 2.00 is approximately 3.60 x 10^(-3).