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The active ingredient in aspirin is acetylsalicylic acid. A \(2.51-\mathrm{g}\) sample of acetylsalicylic acid required \(27.36 \mathrm{mL}\) of \(0.5106 M\) NaOH for complete reaction. Addition of \(13.68 \mathrm{mL}\) of \(0.5106 \mathrm{M}\) 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 \(179.6\,g/mol\), and its Ka value is \(3.311 \times 10^{-4}\).

Step by step solution

01

Find the moles of NaOH that reacted with acetylsalicylic acid

Given that the volume of NaOH is 27.36 mL, and its molarity is 0.5106 M, we can calculate the moles of NaOH that reacted: moles of NaOH = volume × molarity \(moles_{NaOH} = 27.36\times 10^{-3} L \times 0.5106\,M = 0.01397\,mol\)
02

Find the molar mass of acetylsalicylic acid

The moles of acetylsalicylic acid that reacted with NaOH will be equal to the moles of NaOH since they react with a 1:1 stoichiometry. The mass of acetylsalicylic acid is given as 2.51 g. Now we can find the molar mass: Molar mass = mass / moles \(Molar\,mass_{ASA} = \frac{2.51\,g}{0.01397\,mol} = 179.6\,g/mol\)
03

Find the moles of HCl that reacted with the sodium salt

Given that the volume of HCl is 13.68 mL, and its molarity is 0.5106 M, we can calculate the moles of HCl that reacted: moles of HCl = volume × molarity \(moles_{HCl} = 13.68\times 10^{-3} L \times 0.5106\,M = 0.006986\,mol\)
04

Calculate the final concentrations of the species in the mixture

Determine the moles of acetylsalicylic acid (ASA), sodium salt of ASA (NaASA), and HCl that remain after the reaction: \(Moles_{ASA} = 0\,mol\) \(Moles_{NaASA} = moles_{NaOH} - moles_{HCl} = 0.01397 - 0.006986 = 0.006984\,mol\) \(Moles_{H^+} = moles_{HCl} = 0.006986\,mol\) Calculate the final volume of the mixture: Total volume = volume of NaOH + volume of HCl Total volume = 27.36 mL + 13.68 mL = 41.04 mL Total volume = 41.04 mL × \(\frac{1\,L}{1000\,mL}\) = 0.04104 L Using the final volume of the mixture, calculate the final concentrations of the species: \[ [NaASA] = \frac{moles_{NaASA}}{total\,volume} =\frac{0.006984\,mol}{0.04104\,L} = 0.1701\,M\] \[ [H^+] = \frac{moles_{H^+}}{total\,volume} = \frac{0.006986\,mol}{0.04104\,L} = 0.1702\,M\]
05

Calculate the Ka value of acetylsalicylic acid using the pH value

The pH of the solution is given as 3.48. Now we can find the Ka value: \[Ka = \frac{[H^+][ASA^-]}{[ASA]}\] Since [H+] and [ASA] are equal in the final solution, the equation becomes: \[Ka = [H^+]\] To find [H+], use the given pH value: \[pH = -\log[H^+]\] \[H^+ = 10^{-pH} = 10^{-3.48} = 3.311 \times 10^{-4}\] Therefore, the Ka value for acetylsalicylic acid is: \[Ka = 3.311 \times 10^{-4}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Understanding the molar mass of a compound is a fundamental concept in chemistry that allows us to translate between the mass of a substance and the number of moles it contains. For the active ingredient in aspirin, acetylsalicylic acid, this conversion is crucial when analyzing its interactions in a chemical reaction.

To calculate the molar mass, first weigh the substance (in this case, 2.51 grams of acetylsalicylic acid) and then find the number of moles. Since molarity (M) expresses the moles of solute per liter of solution, and the volume of the solution is given in milliliters, it's important to convert the volume to liters by dividing by 1000. The resulting moles of NaOH, which equals the moles of acetylsalicylic acid due to the 1:1 reaction ratio, allows us to use the formula:
\[Molar\, mass = \frac{mass}{moles}\]
With the measured mass and calculated moles, this straightforward division provides the molar mass of acetylsalicylic acid.
Acid Dissociation Constant (Ka)
The acid dissociation constant, or Ka, is a quantitative measure of the strength of an acid in solution. It represents the equilibrium constant for the dissociation of the acid into its conjugate base and a hydrogen ion. The smaller the Ka, the weaker the acid and the less it dissociates.

For acetylsalicylic acid, determining the Ka involves understanding how it reacts with a strong base like NaOH and then back reacts with a strong acid such as HCl. By firstly calculating the moles of NaOH and HCl involved, we can monitor the concentration of the hydrogen ions (H+). Given the pH of the final mixture, we can deduce the concentration of hydrogen ions:\[H^+ = 10^{-pH}\]
With these concentrations, the Ka of the acid becomes efficiently calculable by setting it equal to the concentration of H+ ions since the concentration of acetylsalicylic acid and its conjugate base are equal. Thus, the Ka value can be directly inferred as indicative of the acid’s strength in solution.
Stoichiometry
Stoichiometry is the portion of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It’s based on the conservation of mass where the total mass of the reactants equals the total mass of the products. In the reaction between acetylsalicylic acid and NaOH, we use stoichiometry to determine the amount of NaOH needed to fully react with the given mass of acetylsalicylic acid in a 1:1 ratio.

The balanced chemical reaction stipulates that one molecule of acetylsalicylic acid reacts with one molecule of NaOH, which is why the moles of the acid and the base are equal. Later, when HCl is added to the reaction vessel, stoichiometry again helps us track the changes in the compounds' mole numbers. Knowing that the moles of acid and base must remain balanced, we can determine the concentrations of the various species in the mixture and hence gather details required for calculating the Ka value. Stoichiometry thus serves as the backbone for quantitative analysis in chemical reaction processes.

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Most popular questions from this chapter

A student dissolves 0.0100 mole 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.

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH - to form A Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right)\). Thus how can we claim that a buffered solution resists changes in the pH of the solution?" How would you explain buffering to this friend?

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCN by \(0.100 M \mathrm{KOH}\) at \(25^{\circ} \mathrm{C} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN}=6.2 \times 10^{-10} .\right)\) a. Calculate the \(\mathrm{pH}\) after \(0.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. b. Calculate the \(\mathrm{pH}\) after \(50.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. c. Calculate the \(\mathrm{pH}\) after \(75.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 125 mL of KOH has been added.

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\).

Consider \(1.0 \mathrm{L}\) of a solution that is \(0.85 \mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\) and \(0.80 M \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}} \text { for } \mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} .\right)\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after 0.10 mole of HCl has been added to the original solution. Assume no volume change on addition of HCl. c. Calculate the \(\mathrm{pH}\) after 0.20 mole of \(\mathrm{NaOH}\) has been added to the original buffer solution. Assume no volume change on addition of NaOH.

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