91Ó°ÊÓ

Consider the following two acids: O=C(O)c1ccccc1O Salicylic acid $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ Adipic acid $\mathrm{p} K_{\mathrm{a}_{\mathrm{l}}}=4.4 \mathrm{l} ; \mathrm{p} K_{\mathrm{L}_{3}}=5.28$ In two separate experiments the \(\mathrm{pH}\) was measured during the titration of \(5.00 \mathrm{mmol}\) of each acid with $0.200 \mathrm{M} \mathrm{NaOH}$. Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at $25.00 \mathrm{~mL}\( added \)\mathrm{NaOH}$, and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise 103.)

Step by step solution

01

Identifying the monoprotic and diprotic acid

The first step in understanding these results is to recognize that Salicylic acid is monoprotic (having one ionizable hydrogen), and Adipic acid is diprotic (having two ionizable hydrogens). Adipic acid also has given pKa values, indicating its two dissociable protons.

02

Calculate the amount of NaOH needed to reach the stoichiometric point for each acid

To find the stoichiometric point of a reaction, we can use the formula: moles of acid = moles of base First, let's calculate the moles of each acid to be titrated: For both acids: moles = 5.00 mmol Now, we can calculate the moles of NaOH needed to reach the stoichiometric point for each acid: For Salicylic acid (monoprotic): moles of NaOH = (1 moles of NaOH / 1 moles of acid) × 5.00 mmol = 5.00 mmol For Adipic acid (diprotic): moles of NaOH = (2 moles of NaOH / 1 moles of acid) × 5.00 mmol = 10.00 mmol

03

Calculate the volume of NaOH required for the stoichiometric point of each reaction

Now we can use the moles and the concentration of NaOH to calculate the volume of NaOH required to reach the stoichiometric point for each reaction: Concentration * Volume = moles For Salicylic acid: 0.200 M * Volume = 5.00 mmol Volume = \( \frac{5.00}{0.200} \) mL = 25.00 mL For Adipic acid: 0.200 M * Volume = 10.00 mmol Volume = \( \frac{10.00}{0.200} \) mL = 50.00 mL

04

Explain the titration results based on stoichiometric points

Since our calculations show that Salicylic acid requires 25.00 mL of NaOH, and Adipic acid requires 50.00 mL of NaOH to reach the stoichiometric point, we can now understand the experimental results: The experiment with the stoichiometric point at 25.00 mL added NaOH refers to the titration of Salicylic acid, a monoprotic acid because it requires equal moles of NaOH as the moles of the acid. The experiment with the stoichiometric point at 50.00 mL added NaOH refers to the titration of Adipic acid, a diprotic acid because it requires twice as many moles of NaOH as the moles of the acid to reach the stoichiometric point.

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