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Chapter 15: Problem 104

A 0.210 -g sample of an acid (molar mass \(=192 \mathrm{g} / \mathrm{mol}\) ) is titrated with 30.5 \(\mathrm{mL}\) of 0.108\(M \mathrm{NaOH}\) to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

Short Answer

Expert verified
The acid is likely triprotic, as the calculated ratio of moles of NaOH to moles of the acid is approximately 3, indicating that it has three acidic protons.

Step by step solution

01

Calculate moles of the acid

To find the moles of the acid, we can use the formula: moles = mass / molar mass. We are given the mass (0.210 g) and molar mass (192 g/mol) of the acid. Let's calculate the moles of the acid: Moles of the acid = \(\frac{0.210 \,\text{g}}{192\, \frac{\text{g}}{\text{mol}}}\) Moles of the acid = \(1.094 \times 10^{-3}\) mol
02

Calculate moles of NaOH

To find the moles of NaOH, we can use the formula: moles = volume × concentration. We are given the volume (30.5 mL = 0.0305 L) of NaOH and its concentration (0.108 M). Let's calculate the moles of NaOH: Moles of NaOH = (0.0305 L) × (0.108 mol/L) Moles of NaOH = 0.00329 mol
03

Calculate the ratio of moles of NaOH to molesh

Now, we will calculate the ratio of moles of NaOH to moles of the acid: Ratio = \(\frac{\text{moles of NaOH}}{\text{moles of the acid}}\) Ratio = \(\frac{0.00329\, \text{mol}}{1.094 \times 10^{-3}\, \text{mol}}\) Ratio = 3.005
04

Determine the number of acidic protons

The ratio of moles of NaOH to moles of the acid is approximately 3. Since the ratio is close to 3, we can conclude that the acid is likely triprotic, meaning it has three acidic protons.

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Most popular questions from this chapter

You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the pH of the solution, and how would you use this information? How does the pH of the solution of just the HA compare with that of the final mixture? Explain

You have 75.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) HA. After adding 30.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{NaOH}\) , the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA}\) ?

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{1}}=5.28 $$ In two separate experiments the 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 NaOH, and in the other experiment the stoichiometric point was at 50.00 \(\mathrm{mL}\) NaOH. Explain these results.

One method for determining the purity of aspirin $\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)$ is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) $$ \mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ A sample of aspirin with a mass of 1.427 g was boiled in 50.00 \(\mathrm{mL}\) of 0.500\(M \mathrm{NaOH}\) . After the solution was cooled, it took 31.92 \(\mathrm{mL}\) of 0.289 \(\mathrm{M} \mathrm{HCl}\) to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by 0.100 $\mathrm{M} \mathrm{KOH}\( . Calculate the \)\mathrm{pH}$ of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 150.0 \mathrm{mL}} \\ {\text { b. } 50.0 \mathrm{mL}} & {\text { e. } 200.0 \mathrm{mL}} \\ {\text { c. } 100.0 \mathrm{mL}} & {\text { f. } 250.0 \mathrm{mL}}\end{array} $$
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