The active ingredient in aspirin is acetylsalicylic acid. A 2.51 -g sample of acetylsalicylic acid required 27.36 \(\mathrm{mL}\) of 0.5106 $\mathrm{M} \mathrm{daOH}\( for complete reaction. Addition of 13.68 \)\mathrm{mL}$ of 0.5106\(M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH \(=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{2}\) value. State any assumptions you must make to reach your answer.

Since we have the volume and molar concentration of \(\mathrm{daOH}\) used in the reaction, we can determine the moles of \(\mathrm{daOH}\) used in the reaction. Moles can be calculated using the formula: Moles = Volume (L) × Molarity (M) The volume of NaOH used is given as 27.36 mL, which is equivalent to 0.02736 L (since 1000 mL = 1 L). The molarity of NaOH is given as 0.5106 M. Now, we can calculate the moles of NaOH: Moles of \(\mathrm{daOH}\) = 0.02736 L × 0.5106 M = 0.01396 moles

The moles of acetylsalicylic acid used in the reaction will be equal to the moles of \(\mathrm{daOH}\) used. This is because both react in a 1:1 ratio according to the balanced reaction equation. Therefore the moles of acetylsalicylic acid used are also 0.01396 moles.

The molar mass of a substance can be found using the formula: Molar Mass = mass (g) / moles For acetylsalicylic acid, we are given a mass of 2.51 g and we’ve already calculated the moles of the acid (0.01396 moles). So the molar mass of acetylsalicylic acid is: Molar Mass = 2.51 g / 0.01396 moles = 179.8 g/mol

We are given the volume and molarity of HCl added to the flask (13.68 mL and 0.5106 M, respectively). The HCl reacts with the remaining \(\mathrm{OH}^-\) ions in a 1:1 ratio according to the reaction equation. First, we need to calculate the moles of HCl added: Moles of \(\mathrm{HCl}\) = 0.01368 L × 0.5106 M = 0.006983 moles We can now find the moles of remaining \(\mathrm{OH}^-\) ions by subtracting the moles of HCl from the initial moles of OH-: Moles of remaining \(\mathrm{OH}^-\) ions = Moles of \(\mathrm{daOH}\) - Moles of \(\mathrm{HCl}\) = 0.01396 moles - 0.006983 moles = 0.006977 moles

After adding the HCl to the reaction mixture, the pH of the mixture is given as 3.48. We can use the pH value to find the concentration of \(H^+\) ions in the mixture: \([H^+] = 10^{-\mathrm{pH}} = 10^{-3.48} = 3.31 × 10^{-4}\: \mathrm{M}\) We can now calculate the concentration of the remaining \(\mathrm{OH}^-\) ions: \([\mathrm{OH}^-] = \frac{\text{moles of remaining }\mathrm{OH}^-}{\text{total volume of the reaction mixture (in L)}}\) Total volume of the reaction mixture = Volume of NaOH + Volume of HCl = 0.02736 L + 0.01368 L = 0.04104 L So the concentration of remaining \(\mathrm{OH}^-\) ions is: \([\mathrm{OH}^-] = \frac{0.006977\: \mathrm{moles}}{0.04104\: \mathrm{L}} = 0.170\: \mathrm{M}\) Now we can use the \(K_w\) equation to determine the \(K_w\) value for this reaction: \(K_w = [H^+] × [\mathrm{OH}^-]\) Assuming a value of \(1 × 10^{-14}\) for \(K_w\) at 25°C, we can find the \(K_2\) value: \(K_2 = \frac{K_w}{[H^+]} = \frac{1 × 10^{-14}}{3.31 × 10^{-4}\: \mathrm{M}} = 3.02 × 10^{-11}\) So the molar mass of acetylsalicylic acid is 179.8 g/mol and its \(K_2\) value is \(3.02 × 10^{-11}\).