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Chapter 17: Problem 109

(a) A 0.1044 -g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

Short Answer

Expert verified
(a) The molecular weight of the unknown acid is approximately 94.39 g/mol. (b) The Ka of the acid is approximately 1.29 x 10^{-5}. (c) The unknown monoprotic acid is likely Propanoic Acid, as both its molecular weight (94.12 g/mol) and Ka (1.3 x 10^{-5}) agree with our calculated values.

Step by step solution

01

(a) Calculate the molecular weight

For this step, we will use the data from the titration of the acid and base: - Mass of the acid: 0.1044 g - Volume of NaOH solution to reach the end point: 22.10 mL - Concentration of the NaOH solution: 0.0500 M Since it is a monoprotic acid, one mole of acid reacts with one mole of base (NaOH). 1. Calculate the number of moles of NaOH by using the volume and concentration: \(Moles\ of\ NaOH = Volume\ of\ NaOH\ ×\ Concentration\ of\ NaOH\) \(Moles\ of\ NaOH = 0.0221 L × 0.0500 M = 0.001105\ moles\) 2. Determine the moles of the unknown acid: Since one mole of acid reacts with one mole of base (NaOH), the moles of the unknown acid equal the moles of NaOH: Moles of acid = 0.001105 moles 3. Calculate the molecular weight of the acid: \(Molecular\ Weight\ of\ Acid = \frac{Mass\ of\ Acid}{Moles\ of\ Acid}\) Molecular Weight of Acid = \(\frac{0.1044\ g}{0.001105\ moles} ≈ 94.39\ g/mol\)
02

(b) Calculate the Ka of the acid

We know that at half-equivalence point, the pH of the solution equals the pKa of the acid. The pH after titrating half of the required base volume (11.05 mL) is 4.89. Therefore, we can directly calculate the Ka of the acid: 1. Find the pKa of the acid: pKa of Acid = pH at half-equivalence point = 4.89 2. Calculate the Ka of the acid: Ka = \(10^{-pKa}\) Ka ≈ \(10^{-4.89} ≈ 1.29 × 10^{-5}\)
03

(c) Identify the acid

To identify the acid, we can use the molecular weight and Ka that we calculated in steps (a) and (b). Comparing these values with those in Appendix D or any other similar source, we can identify the acid whose values are closest to our calculations. After a quick search, the acid that best fits the calculated values is Propanoic Acid: - Molecular weight: 94.12 g/mol - Ka = 1.3 x 10^-5 Our calculated molecular weight and Ka approximately match those of Propanoic Acid. So, the unknown monoprotic acid is likely Propanoic Acid. Both molecular weight and Ka agree with this choice.

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Key Concepts

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

Monoprotic Acid
Understanding the chemistry behind a monoprotic acid is essential for solving titration problems effectively. A monoprotic acid is a type of acid that has only one proton (hydrogen ion) that it can donate to a base during a chemical reaction. The classic example is Hydrochloric acid (\(HCl\)), which dissociates in water to give one proton and one chloride ion (\(Cl^-\)).

This property simplifies titration calculations because the stoichiometry is a one-to-one ratio between the acid and the base used for titration. In the given exercise, the monoprotic nature of the acid allowed us to directly equate the moles of acid with the moles of sodium hydroxide (NaOH) used, simplifying the calculation for molecular weight of the unknown acid.
Molecular Weight Calculation
The molecular weight of a compound represents the mass of one mole of that substance. The unit for expressing molecular weight is grams per mole (g/mol). To calculate the molecular weight of a substance, we need the mass of a sample of the substance and the number of moles in that sample.

In the context of the given problem, we first determined the number of moles of NaOH present in the titration, and then used the stoichiometry of the reaction (assuming a 1:1 molar ratio for monoprotic acid and NaOH) to find the moles of the unknown acid. Dividing the mass of the unknown acid by the moles gave us its molecular weight. This process is fundamental in analytical chemistry, as it allows us to deduce properties about an unknown substance through experimental means.
Acid Dissociation Constant (Ka)
The acid dissociation constant, Ka, is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation reaction of the acid into its ions. A higher Ka value indicates a stronger acid, which implies it dissociates more completely in water.

In the given exercise, we calculated the Ka using the pH value measured at the half-equivalence point in a titration. At this point, the concentration of the acid equals the concentration of its conjugate base, simplifying the expression for Ka and allowing us to use the pKa as the negative logarithm of Ka. This relationship demonstrates the interplay between equilibrium chemistry and the acid-base reactions in solutions, helping us identify the strength of the unknown acid.
pH Calculation
pH is a scale used to specify the acidity or basicity of an aqueous solution. It is the negative base 10 logarithm of the molar concentration of hydrogen ions (\(H^+\)) in the solution. A pH less than 7 is acidic, and a pH greater than 7 is basic. Calculating pH is a fundamental part of understanding chemical equilibrium in acid-base reactions.

In titration problems, such as the one given, pH calculations help identify the point at which equal amounts of acid and base have reacted. Halfway to the equivalence point, the pH equals the pKa, which is derived from the Ka. This knowledge enables students to understand the importance of pH in determining the properties of acids and the progression of titration reactions.

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Most popular questions from this chapter

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH} 7:\) (a) formic acid titrated with \(\mathrm{NaOH},\) (b) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid.

A 20.0 -mL sample of \(0.150 \mathrm{M} \mathrm{KOH}\) is titrated with \(0.125 \mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{~mL},\) (b) \(23.0 \mathrm{~mL},\) (c) \(24.0 \mathrm{~mL}\), (d) \(25.0 \mathrm{~mL},(\mathrm{e}) 30.0 \mathrm{~mL}\).

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: AgI, \(\mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2},\) and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\)

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}, 0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\) and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction \(\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) (c) If we assume that the only sources of \(\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-},\) and \(\mathrm{OH}^{-}\) ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the preceding expression? What is the \(\mathrm{pH} ?\) (d) If the \(\mathrm{pH}\) is buffered at 8.2 (as is historically typical for the ocean), what is the molar solubility of \(\mathrm{CaCO}_{3} ?\) (e) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\) How much does this drop in \(\mathrm{pH}\) increase solubility? solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.
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