presentation Tasks:

 Ephedrine (structure see below) is a natural product that is used as a medicine in the treatment of asthma. The corresponding acid to ephedrine have pKa = 9.60 at 25 ° C.
  Question:
 Calculate the pH reaches 991 mg ephedrine dissolved in 50.0 mL of water at 25 ° C.

To calculate the pH of a solution containing ephedrine, we need to consider its acid-base properties. Ephedrine has a corresponding acid with a pKa value of 9.60.

At the given conditions, we have 991 mg of ephedrine dissolved in 50.0 mL of water at 25 °C. We need to convert the mass of ephedrine to moles to make further calculations.

The molar mass of ephedrine is calculated by summing the atomic masses of all the atoms in the molecule, which are as follows:
C: 12.01 g/mol
H: 1.01 g/mol
N: 14.01 g/mol

The molar mass of ephedrine is: (10 x 12.01 g/mol) + (15 x 1.01 g/mol) + 1 x 14.01 g/mol = 150.21 g/mol

Using the molar mass, we can calculate the number of moles of ephedrine:

moles of ephedrine = mass of ephedrine / molar mass of ephedrine
= 991 mg / 150.21 g/mol
= 0.0066 moles

Since the volume of the solution is given as 50.0 mL, we can convert it to liters:

volume of solution = 50.0 mL = 0.0500 L

Now, we can calculate the molarity (M) of the ephedrine solution by dividing the number of moles by the volume of the solution in liters:

Molarity (M) = moles of ephedrine / volume of solution
= 0.0066 moles / 0.0500 L
= 0.132 M

Given that the pKa of the acid form of ephedrine is 9.60, we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log ([A-] / [HA])

Where:
pH = the pH of the solution
pKa = the pKa of the acid form of ephedrine
[A-] = concentration of the conjugate base (ephedrine ionized)
[HA] = concentration of the acid form (ephedrine protonated)

Since the pH is a measure of the concentration of H+ ions, and ephedrine acts as a base, we need to calculate the concentration of the conjugate acid form, [HA]:

[HA] = concentration of ephedrine - [A-]
= 0.132 M - [A-]

As ephedrine acts as a weak base, most of the ephedrine will be protonated in water, and the concentration of A- is negligible compared to the concentration of HA. Therefore, we can ignore [A-]:

[HA] ≈ 0.132 M

Substituting the values into the Henderson-Hasselbalch equation:

pH = 9.60 + log (0.132 M / 0.132 M)
= 9.60 + log (1)
= 9.60

Therefore, the pH of the solution containing 991 mg of ephedrine dissolved in 50.0 mL of water at 25 °C is 9.60.