A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's K_a value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.
To find the pKa of X-281, you prepare a 0.086 \it M test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKa of X-281?
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Your post lists 0.086\it M as the concentration . Is that 0.086 M or jus what? I'll assume it is and you can adjust the answer below if it is something other than that.
Call the new acid HX, then
HX ==> H^+ + X^-
Write the Ka expression.
Ka = (H^+)(X^-)/(HX)
You know pH which you can convert to (H^+). (X^-) must be the same. (HX) = 0.086-(H^+). Solve for Ka and convert to pKa. Post your work if you get stuck.
4.32
To find the pKa of X-281 based on the provided information, we need to make use of the Henderson-Hasselbalch equation, which relates the pH, pKa, and the ratio of the concentration of the conjugate base to the concentration of the acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the measured pH of the solution (2.70 in this case)
- pKa is the negative logarithm of the acid dissociation constant (unknown)
- [A-] is the concentration of the conjugate base (undissociated form)
- [HA] is the concentration of the acid (dissociated form)
In this case, X-281 is a weak acid, so we can assume that most of the acid will dissociate into [A-] and H+ ions. Therefore, we can approximate [A-] to be equal to the initial concentration of X-281 in the solution (0.086 M).
Now we can rearrange the Henderson-Hasselbalch equation to solve for the pKa:
pKa = pH - log([A-]/[HA])
In this case, we can assume that the concentration of [A-] is equal to the initial concentration of X-281 (0.086 M) since most of it dissociates.
Let's substitute the given values into the equation and solve:
pKa = 2.70 - log(0.086/0.086)
pKa = 2.70 - log(1)
pKa = 2.70 - 0
pKa = 2.70
Therefore, the pKa of X-281 is 2.70.
To find the pKa of X-281, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the dissociated form of the weak acid and [HA] represents the concentration of the undissociated (protonated) form of the weak acid.
In this case, X-281 is a monoprotic weak acid, meaning it can release only one proton (H+).
Given the pH of the solution (2.70), we can assume that the majority of X-281 is in the undissociated form (HA). This is because in an acidic solution (pH less than pKa), the weak acid tends to stay protonated.
We are given that the concentration of the X-281 test solution is 0.086 M. Since we are assuming most of it is HA, we can assume [HA] = 0.086 M.
Now we can rearrange the Henderson-Hasselbalch equation to solve for pKa:
pKa = pH - log([A-]/[HA])
To find [A-], we can use the fact that the total concentration of X-281 is the sum of the concentrations of HA and A-. Since A- and HA represent the dissociated and undissociated forms respectively, we have:
[A-] + [HA] = 0.086 M
Since we assumed that most of the X-281 is in the undissociated form, we can also assume that [A-] is negligible compared to [HA]. This means we can approximate [A-] as 0 in the equation above.
Thus, we have:
0 + [HA] = 0.086 M
Solving for [HA]:
[HA] = 0.086 M
Now we can substitute the values into the Henderson-Hasselbalch equation:
pKa = 2.70 - log(0/0.086)
Since log(0/0.086) is undefined, we can determine that the X-281 solution is more acidic than the calculated pH of 2.70. This implies that the majority of X-281 is undissociated, with only a small fraction being in the dissociated form.
This means that the pKa of X-281 is less than 2.70 but cannot be precisely determined with the given information.