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Chapter 15: Problem 79

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

Short Answer

Expert verified
The molar mass of acetylsalicylic acid is approximately \(180.16 \: \mathrm{g/mol}\) and its \(K_a\) value is approximately \(2.97 \times 10^{-4}\). We have assumed ideal behaviors, equal stoichiometry between the reactants, and that the reaction is fully completed as described in the problem.

Step by step solution

01

Find the moles of NaOH and HCl

Calculate the moles of \(\mathrm{NaOH}\) and \(\mathrm{HCl}\) used in the reactions: $$\text{moles of NaOH} = \text{Volume} \times \text{Molarity} = 27.36\: \text{mL} \times 0.5106\: \text{M}$$ $$\text{moles of HCl} = \text{Volume} \times \text{Molarity} = 13.68\: \text{mL} \times 0.5106\: \text{M}$$
02

Calculate the moles of acetylsalicylic acid

Since one mole of acetylsalicylic acid reacts with one mole of NaOH, the moles of acetylsalicylic acid present in the sample can be equated to the moles of NaOH: $$\text{moles of acetylsalicylic acid} = \text{moles of NaOH}$$
03

Calculate the molar mass of acetylsalicylic acid

Divide the mass of the acetylsalicylic acid sample by the moles of acetylsalicylic acid to determine its molar mass: $$\text{Molar mass of acetylsalicylic acid} = \frac{\text{Mass}}{\text{Moles}} = \frac{2.51\: \text{g}}{\text{moles of acetylsalicylic acid}}$$
04

Calculate the moles of remaining acetylsalicylate ions

Acetylsalicylate ions are the conjugate bases of acetylsalicylic acid. After adding HCl, some of these ions will react with the HCl to form acetylsalicylic acid again. Find the remaining moles of acetylsalicylate ions by subtracting the moles of HCl from the total moles of acetylsalicylic acid. $$\text{moles of acetylsalicylate ions} = \text{moles of acetylsalicylic acid - moles of HCl}$$
05

Calculate the concentration of acetylsalicylate ions

Divide the moles of acetylsalicylate ions by the total volume of the mixture to determine their concentration: $$\text{Concentration of acetylsalicylate ions} = \frac{\text{Moles of acetylsalicylate ions}}{\text{Total volume}}$$ The total volume is the sum of the volumes of the three solutions, in this case, the NaOH, HCl, and acetylsalicylic acid solutions.
06

Calculate the pH and [H+] concentration in the mixture

Now that we know the concentration of acetylsalicylate ions, we can use the given pH to calculate the concentration of H+ ions in the mixture. $$\mathrm{pH} = -\log{[\mathrm{H^+}]}$$ Solving for \([\mathrm{H^+}]\): $$[\mathrm{H^+}] = 10^{-\mathrm{pH}}$$
07

Calculate Ka of acetylsalicylic acid

Finally, we can use the values of acetylsalicylic acid, its conjugate base, and \([\mathrm{H^+}]\) to find the \(K_a\): $$K_a = \frac{[\mathrm{H^+}][\mathrm{acetylsalicylate}]}{[\mathrm{acetylsalicylic\:acid}]}$$ With this, we have determined the molar mass of acetylsalicylic acid and its \(K_a\) value. Any assumptions made along the way should be stated, including any ideal behaviors or assumptions about the reaction stoichiometry.

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Most popular questions from this chapter

You make \(1.00 \mathrm{~L}\) of a buffered solution \((\mathrm{pH}=4.00)\) by mixing acetic acid and sodium acetate. You have \(1.00 M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

A certain acetic acid solution has \(\mathrm{pH}=2.68\). Calculate the volume of \(0.0975 M \mathrm{KOH}\) required to reach the equivalence point in the titration of \(25.0 \mathrm{~mL}\) of the acetic acid solution.

Consider a solution containing \(0.10 M\) ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right.\) ), \(0.20 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{3}^{+}\), and \(0.20 \mathrm{M} \mathrm{Cl}^{-}\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after \(0.050 \mathrm{~mol} \mathrm{KOH}(s)\) is added to \(1.00 \mathrm{~L}\) of this solution. (Ignore any volume changes.)

A certain indicator HIn has a \(\mathrm{p} K_{2}\) of \(3.00\) and a color change becomes visible when \(7.00 \%\) of the indicator has been converted to \(\mathrm{In}^{-}\). At what \(\mathrm{pH}\) is this color change visible?

A student dissolves \(0.0100 \mathrm{~mol}\) of an unknown weak base in \(100.0 \mathrm{~mL}\) water and titrates the solution with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\). After \(40.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.
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