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Malonic acid \(\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) is a diprotic acid. In the titration of malonic acid with NaOH, stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8 . \mathrm{A} 25.00-\mathrm{mL}\) sample of malonic acid of unknown concentration is titrated with 0.0984 \(\mathrm{M}\) NaOH, requiring 31.50 \(\mathrm{mL}\) of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution.

Short Answer

Expert verified
The concentration of the initial malonic acid solution is approximately 0.12374 M.

Step by step solution

01

Write the balanced chemical equation for the second ionization reaction of malonic acid

At the second stoichiometric point, both protons of malonic acid are neutralized by the \(\mathrm{NaOH}\). The balanced chemical equation for the second ionization reaction is given by: \(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2}^{2-} + \mathrm{H}_{2} \mathrm{O}\)
02

Find moles of NaOH

Next, we need to find the moles of \(\mathrm{NaOH}\) used in the reaction. Use the formula: moles = concentration × volume; where volume should be in liters. Moles of NaOH = (0.0984 mol/L) × (31.50 mL × (1 L / 1000 mL)) Moles of NaOH = 0.0030936 mol
03

Find moles of malonic acid

Since the stoichiometry between the \(\mathrm{NaOH}\) and malonic acid is 1:1 in the second ionization reaction, the moles of malonic acid required are equal to the moles of \(\mathrm{NaOH}\) used. Moles of malonic acid = 0.0030936 mol
04

Calculate the concentration of malonic acid

Now, we'll use the initial volume of the malonic acid solution and the moles of malonic acid to find its concentration. Use the formula: concentration = moles / volume; where volume should be in liters. Concentration of malonic acid = (0.0030936 mol) / (25.00 mL × (1 L / 1000 mL)) Concentration of malonic acid = 0.12374 M Hence, the concentration of the initial malonic acid solution is approximately 0.12374 M.

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Key Concepts

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

Diprotic Acid
A diprotic acid is a type of acid that can donate two protons (hydrogen ions) during its dissociation process in an aqueous solution. Diprotic acids undergo ionization in two steps rather than all at once. The first ionization releases the first proton and forms a conjugate base. The second ionization subsequently releases the second proton from the conjugate base. Each step has its unique ionization constant, indicating how readily the protons are released during each step. Malonic acid, for example, is a diprotic acid with two carboxylic groups, which allows it to donate two protons. In the chemical task at hand, understanding the behavior of diprotic acids is crucial, as different protons are removed during the two distinct stages of the titration.
Chemical Stoichiometry
Chemical stoichiometry refers to the quantitative relationships between the reactants and products in a balanced chemical equation. It relates to mole ratios and is grounded on the balanced chemical equation principle. Stoichiometry is fundamental when performing titrations because it ensures that reactants are proportioned correctly to fully react without any leftover substances. In the original exercise, the stoichiometric relationship between malonic acid and sodium hydroxide (\( \text{NaOH} \)) is established by balancing the equations, enabling us to calculate the unknown concentration. By knowing the precise proportions of each reactant, we can determine how much of an acid reacts with a given volume and concentration of a base.
Acid-Base Titration
Acid-base titration is a laboratory technique that determines the concentration of an acid or base in a solution. During titration, a reagent of known concentration, called the titrant (such as NaOH), is gradually added to the analyte until a reaction between equivalent amounts has occurred, indicated by a change in color with an indicator. This point is known as the endpoint. In diprotic acids like malonic acid, there are two distinct endpoints for each proton being neutralized. Observing and calculating the titration precisely allows for determining the unknown concentration of an acid, providing valuable insights into the chemical behavior of the substance being analyzed.
Malonic Acid
Malonic acid, structured as \( \text{HO}_2 \text{CCH}_2 \text{CO}_2 \text{H} \), is a dicarboxylic acid known for its two acidic carboxyl groups capable of releasing protons. It is a well-studied diprotic acid and is widely used in organic synthesis methods. In chemical analyses, malonic acid exemplifies the dual ionization process exhibited by diprotic acids. During titrations involving malonic acid and a base like NaOH, understanding its structure and behavior underlies why two stoichiometric points are observed, allowing calculations to determine concentrations precisely. Its versatile properties make malonic acid a pivotal compound for both educational and practical applications in chemistry.
Chemical Ionization
Chemical ionization is a crucial process where molecules gain or lose ions, resulting in ions that are more conducive to analysis and measurement. This phenomenon is particularly significant during titration, where ionization allows for changing the pH and observing endpoints. In the context of organic acids such as malonic acid, ionization reflects how readily hydrogen ions are released in aqueous solutions. Having a clear insight into chemical ionization helps students understand how malonic acid reacts with bases during titration, facilitating the observation of the entire acid dissociation mechanism. Consequently, grasping this concept means better anticipating the behavior of acids and bases in various analytical settings.

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Most popular questions from this chapter

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M}\) \(\mathrm{HCl} .\)

Calculate the pH of a solution formed by mixing 100.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) NaF and 100.0 \(\mathrm{mL}\) of 0.025 \(\mathrm{M} \mathrm{HCl} .\)

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 titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HONH}_{2}\) by 0.100 \(\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)\) a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

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?

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