Question

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 \(\mathrm{NaOH}\), stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8\). A \(25.00-\mathrm{mL}\) sample of malonic acid of unknown concentration is titrated with \(0.0984 \mathrm{M} \mathrm{NaOH}\), requiring \(31.50 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (See Exercise 113.)

Short answer

The concentration of the initial malonic acid solution is approximately \(0.0621 \, M\).

Step-by-step solution

Analyze the titration reaction

Let's first write down the balanced chemical equation for the titration reaction happening at the second stoichiometric point (pH=8.8): \(HO_2CCH_2CO_2H + 2 NaOH\longrightarrow Na_2[H_2O_2CCH_2CO_2]\) At the second stoichiometric point, one molecule of diprotic malonic acid reacts with two molecules of NaOH.

Calculate the moles of NaOH used in the reaction

We know the concentration and volume of NaOH solution used for titration. Using this information, we can calculate the moles of NaOH. Moles of NaOH = Molarity × Volume The volume must be converted to liters: Volume of NaOH = \(31.50 \, mL = 31.50 \times 10^{-3} \, L\) Moles of NaOH = \(0.0984\,M \times 31.50 \times 10^{-3} \, L = 3.10464 \times 10^{-3} \, moles\)

Calculate the moles of malonic acid

In the balanced chemical equation, we see that the one molecule of malonic acid reacts with two molecules of NaOH. Hence, we can simply divide the moles of NaOH by 2 to obtain the moles of malonic acid: Moles of malonic acid = Moles of NaOH / 2 Moles of malonic acid = \(3.10464 \times 10^{-3} \, moles / 2 = 1.55232 \times 10^{-3} \, moles\)

Calculate the concentration of the initial malonic acid solution

Finally, we can find the concentration of the initial malonic acid solution by dividing the moles of malonic acid by its volume: Concentration of Malonic Acid = Moles of Malonic Acid / Volume of Malonic Acid Solution The volume must be converted to liters: Volume of Malonic Acid Solution = \(25.00\, mL = 25.00 \times 10^{-3} L\) Concentration of Malonic Acid = \(1.55232 \times 10^{-3} \, moles / 25.00 \times 10^{-3} \, L = 0.0620928 \, M\) The concentration of the initial malonic acid solution is approximately \(0.0621 \, M\).

Key concepts

Diprotic Acid

Diprotic acids are a unique category of acids that have two ionizable hydrogen atoms per molecule. This means they can undergo two separate ionizations in an acid-base reaction. A well-known example is sulfuric acid \( (H_2SO_4) \), but in the provided exercise, malonic acid \( (HO_2CCH_2CO_2H) \) serves as the diprotic acid.

In the context of titration, diprotic acids react with a base like sodium hydroxide (NaOH) to form neutral water and a salt. This reaction occurs in two steps, as each hydrogen ion reacts separately.

• The first reaction removes the first hydrogen ion, creating a singly ionized anion.
• The second step involves further reaction to remove the second hydrogen ion.

Understanding this dual reaction process is essential for calculations involving stoichiometric points and moles.

Stoichiometric Point

The stoichiometric point in a titration is a crucial concept, representing the moment when the amount of titrant added is chemically equivalent to the substance in the solution being titrated. For diprotic acids like malonic acid, there are typically two stoichiometric points, corresponding to each ionizing hydrogen.

In this exercise, these points occur at pH values of 3.9 and 8.8. These values indicate two milestones in neutralizing each hydrogen atom using a base, where:

• At the first stoichiometric point, only one hydrogen has reacted, resulting in a partially ionized salt.
• At the second stoichiometric point, both hydrogens have reacted, fully neutralizing the acid.

Understanding these points is key to identifying where the endpoints occur in the titration process, aiding in the correct calculation of the original acid concentration.

Moles Calculation

Calculating the number of moles is a vital part of solving titration problems as it provides the quantitative measure of reactants involved. The mole is a fundamental unit in chemistry for counting atomic-scale particles by relating them to a common measurable quantity.

To find the moles of a substance, use the formula:\[ \text{Moles} = \text{Concentration} \times \text{Volume (in liters)} \]For this exercise, we calculated the moles of NaOH used in the titration process:

• Given the molarity (0.0984 M) and converted volume (31.50 mL to 0.03150 L), we determine NaOH moles as \( 3.10464 \times 10^{-3} \).
• Since each molecule of malonic acid reacts with two molecules of NaOH, the moles of malonic acid equal half the moles of NaOH, which is \( 1.55232 \times 10^{-3} \).

Accurate mole calculation is key, as it directly influences the determination of the original concentration of the solution.

Acid-Base Reaction

An acid-base reaction involves the transfer of hydrogen ions (protons) from the acid to the base. These reactions are pivotal in titration processes as they allow the measurement and neutralization of an acid by a known base solution.

In the context of a diprotic acid like malonic acid, this reaction occurs in two phases reflecting the two hydrogen ions available for donation. The letter equation representing the complete neutralization in the case of malonic acid with sodium hydroxide is:\[ \mathrm{HO}_2\mathrm{CCH}_2\mathrm{CO}_2\mathrm{H} + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2[\mathrm{H}_2\mathrm{O}_2\mathrm{CCH}_2\mathrm{CO}_2] + 2\mathrm{H}_2\mathrm{O} \]Understanding the intricacies of acid-base reactions:

• Highlights the stoichiometric relationships during the titration.
• Illustrates how titrations help in determining unknown concentrations through known reactions.

Mastery of these reactions forms the basis of correctly carrying out titrations in a laboratory setting.