CHEM 104 Homework 6 20 points-Ch. 19.1, 19.2 Aspirin is one of the most widely u

Question

CHEM 104 Homework 6 20 points-Ch. 19.1, 19.2 Aspirin is one of the most widely used and versatile medicines, used for treating pain, controlling fever, and preventing heart attacks and blood clots. The active ingredient in aspirin in acetylsalicylic acid. From the experimental information below, determine the molar mass of acetylsalicylic acid and its K, value. State any assumptions you must make to find your answerl 4. A 2.51 g sample of acetylsalicylic acid is titrated exactly with 27.36 mL of 0.5106 M NaOH solution. Addition of 13.68 mL of O.5106 M HCI to the resulting mixture produces a solution with pH 3.48. Data:

Explanation / Answer

Answer:

After titrating Acetylsalicylic acid with NaOH, the mixture titrated with 13.68 mL of 0.5106 M HCl giving pH to the solution as, 3.48.

a) Let us calculate actual [H+] in added to the mixture,

Molarity of HCl = 0.5106 M, Volume = 13.68 mL = 0.01368 L.

Moles of H+ = Molarity * Volume in L = 0.5106 * 0.01368 = 7.0 * 10-3 moles.

b) [H+] left can be calculated from pH

pH = 3.48

[H+] = 10-pH = 10-3.48 = 3.3 * 10-4 M

Tootal volume of the mixture = 27.36 + 13.68 = 41.04 mL = 0.04104 L

Moles of H+ = 3.3*10-4 x 0.04104 = 0.014 * 10-3 moles.

Hence moles of H+ reacted with excess OH- ions are,

Moles of HO- neutralized = Actual H+ ions added - H+ ions left = 7*10-3 - 0.014*10-3 = 6.984 * 10-3 moles.

c) Moles of NaOH i.e. HO- ions

Molarity = 0.5106 M, Volume = 27.36 mL = 0.02736 L

Moles of HO- ions = 0.5106*0.02736 = 14.0 * 10-3 moles.

Hence moles of HO- reacted with Acetylsalicylic acid is,

HO- ions reacted = Actually added - Neutralized by HCl = 14*10-3 - 6.984*10-3 = 7.014*10-3 moles.

Hence,

Moles of Acetylsalicylic acid reacted = Moles of HO- reacted = 7.014*10-3 moles.

Now Mass of Acetylsalicylic acid taken = 2.51 g.

Molar mass of Acetylsalicylic acid = Mass taken in gram / # of Moles = 2.51 / (14.0*10-3) = 179.8 g

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