A laboratory technician wants to determine the aspirin content of a headache pill by acid–base titration. Aspirin has a Ka of 3.0 × 10-4. The pill is dissolved in water to give a solution that is about 10-2 M and is then titrated with KOH solution. Find the pH at each of the following points, neglecting dilution effects:
(a) before titration begins.
(b) at the stoichiometric point.
(c)at the midpoint of the titration.
Hi. I attempted this but got it wrong.. idk why. Can you help me
3.0 × 10-4(10^-2)=x^2
x=3.0E^-5M
x=(10^-2)-3.0E-5M
= 0.09997M
To determine the pH at each of the given points, we need to consider the acidic and basic properties of aspirin and the reaction that occurs during the titration.
(a) Before titration begins:
At this point, we have only the solution of the pill dissolved in water. Aspirin is a weak acid, so we need to calculate the pH by considering the dissociation of aspirin in water using its Ka value.
The dissociation equation of aspirin (HA) in water is:
HA ⇌ H+ + A-
Using the Ka expression: Ka = [H+][A-] / [HA]
Let's assume x mol/L is the concentration of H+ and A- ions formed from the dissociation of aspirin.
Since the initial concentration of aspirin is 10^(-2) M, the initial concentration of HA is also 10^(-2) M.
Plugging in these values into the Ka expression:
3.0 × 10^(-4) = x² / (10^(-2) - x)
Aspirin is a weak acid, so we can assume that x is very small compared to 10^(-2), and thus, 10^(-2) - x ≈ 10^(-2).
3.0 × 10^(-4) = x² / (10^(-2))
Simplifying the equation:
3.0 × 10^(-4) = x² / 10^(-2)
x² = 3.0 × 10^(-6)
x = √(3.0 × 10^(-6)) ≈ 5.48 × 10^(-3)
Since we calculated the concentration of H+ ions, we can calculate the pH using the formula pH = -log[H+]:
pH = -log(5.48 × 10^(-3))
pH ≈ 2.26
So, the pH before titration begins is approximately 2.26.
(b) At the stoichiometric point:
At the stoichiometric point, the number of moles of KOH added is equal to the number of moles of aspirin present.
Since aspirin is a monoprotic acid, it reacts in a 1:1 ratio with the base. This means that at the stoichiometric point, all of the aspirin has reacted with the KOH, and we are left with only the salt formed from the reaction.
The salt formed by the reaction of aspirin with KOH is potassium aspirinate (KA). This salt dissociates to form potassium cations (K+) and aspirinate anions (A-).
Since potassium ions do not affect the pH in this case, we only need to consider the concentration of the aspirinate anions.
At the stoichiometric point, the concentration of aspirinate anions is equal to the initial concentration of aspirin, which is 10^(-2) M.
As potassium aspirinate is a strong electrolyte, it fully dissociates, so the concentration of A- ions is the same as the concentration of potassium aspirinate, which is 10^(-2) M.
Therefore, the pH at the stoichiometric point is determined solely by the concentration of the aspirinate ions. Since we have a 10^(-2) M concentration of A-, the pH can be calculated using the formula pH = -log([H+]):
pH = -log(10^(-2))
pH = 2
So, the pH at the stoichiometric point is 2.
(c) At the midpoint of the titration:
The midpoint of the titration occurs when half of the aspirin has reacted with the KOH. At this point, the concentration of the weak acid and its conjugate base (aspirin and aspirinate) will be equal.
To find the pH at the midpoint, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base.
pH = pKa + log([A-] / [HA])
At the midpoint, the concentration of aspirin (HA) and aspirinate (A-) will be equal. So, the ratio [A-] / [HA] will be 1.
Using the pKa value of 3.0 × 10^(-4) and plugging in the values:
pH = -log(3.0 × 10^(-4)) + log(1)
pH ≈ 3.52
Therefore, the pH at the midpoint of the titration is approximately 3.52.
Make sure to double-check your calculations and check the units used in the calculations. Also, ensure that you consider any simplifications made during the calculations.
To find the pH at each of the given points in the acid-base titration, we need to consider the dissociation of aspirin (acetylsalicylic acid) in water and the subsequent reaction with KOH. Let's proceed step-by-step:
(a) Before titration begins:
At this point, we have only acetylsalicylic acid in solution. Since we neglect dilution effects, the concentration of the acid remains the same as the initial concentration, which is 10^-2 M. To calculate the pH, we need to find the concentration of H+ ions. However, acetylsalicylic acid is a weak acid, so we will need to consider its Ka value.
The Ka expression for acetylsalicylic acid can be written as follows:
Ka = [H+][C9H8O4] / [C9H7O4-]
For acetylsalicylic acid, [H+] is equal to x, [C9H8O4] is equal to (10^-2 - x), and [C9H7O4-] is equal to x.
Plugging in the values, we can write the Ka expression as:
3.0 × 10^-4 = x^2 / (10^-2 - x)
Since x is small compared to 10^-2, we can approximate (10^-2 - x) as 10^-2. Thus, the Ka expression becomes:
3.0 × 10^-4 = x^2 / 10^-2
Rearranging the equation, we find:
x^2 = (3.0 × 10^-4)(10^-2)
x^2 = 3.0 × 10^-6
Taking the square root of both sides, we get:
x = √(3.0 × 10^-6)
x ≈ 5.48 × 10^-4
Since x represents [H+], we can find the pH using the equation:
pH = -log[H+]
pH = -log(5.48 × 10^-4)
pH ≈ 3.26
Therefore, the pH before titration begins is approximately 3.26.
(b) At the stoichiometric point:
At the stoichiometric point, the number of moles of KOH added is equal to the number of moles of acetylsalicylic acid initially present.
Since the acid is monoprotic, the number of moles of acetylsalicylic acid is simply the initial concentration (10^-2 M) multiplied by the volume of the solution.
The stoichiometric point occurs when all of the acetylsalicylic acid is neutralized by KOH, forming water and a salt. In this case, the salt formed will be potassium acetylsalicylate (K+C9H7O4-).
At the stoichiometric point, there will be no excess acetylsalicylic acid or KOH remaining. Therefore, the resulting solution will contain only potassium acetylsalicylate, which is a salt and fully dissociated.
Since a salt is a neutral species, the pH at the stoichiometric point will be approximately 7 (neutral).
(c) At the midpoint of the titration:
The midpoint of the titration occurs when exactly half of the acetylsalicylic acid has been neutralized by KOH.
At this point, the number of moles of acetylsalicylic acid remaining is equal to half of the initial moles, and the number of moles of KOH added is equal to the number of moles of acetylsalicylic acid initially present.
Using the same reasoning as in part (b), we know that the stoichiometric point occurs when 5.48 × 10^-4 moles (or half of the initial moles) of acetylsalicylic acid remain.
Thus, the concentration of acetylsalicylic acid at the midpoint of the titration is:
Concentration = (moles remaining) / (volume of solution)
Concentration = (5.48 × 10^-4 moles) / (volume of solution)
Since we neglected dilution effects, the volume of the solution remains the same as the initial volume.
The pH at the midpoint can be calculated in a similar manner as in part (a), using the concentration of acetylsalicylic acid at the midpoint. Simply follow the steps in part (a), using the new concentration value.
I hope this explanation helps! Let me know if you have any further questions.