Problem 57 - General Chemistry: Principles and Modern Applications

Question

The pH of a solution of \(19.5 \mathrm{g}\) of malonic acid in \(0.250 \mathrm{L}\) is \(1.47 .\) The pH of a \(0.300 \mathrm{M}\) solution of sodium hydrogen malonate is 4.26. What are the values of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for malonic acid?

Step-by-Step Solution

Answer

In order to find the values of \(K_{a_{1}}\) and \(K_{a_{2}}\), first, calculate the concentration of H+ ions from the given pH values. Then, calculate the molarity of the malonic acid using the weight and volume of the solution. Finally, use these values to calculate \(K_{a_{1}}\) and \(K_{a_{2}}\) using their respective formulas. The values will depend upon the specific numerical values obtained during the calculations.

Step 1: Identifying the Given Information

Identify all the given data and write them down. The weight of the malonic acid is given to be 19.5g, volume of the solution: 0.250L, pH of the solution: 1.47. Furthermore, pH of 0.300M solution of sodium hydrogen malonate is 4.26.

Step 2: Using the Formula for pH

For each solution, find the concentration of H+ ions (\([H^+]\)) using the formula \(pH = -log [H^+]\). For solution 1, \( [H^+] = 10^{-pH} = 10^{-1.47} \). For solution 2, \([H^+] = 10^{-pH} = 10^{-4.26}\)

Step 3: Determine the Molar Mass of Malonic Acid

Using the periodic table, find the molar mass of malonic acid(C3H4O4) to be approximately 104g/mol.

Step 4: Find the Concentration of the Malonic Acid

Using the formula \(Molarity = \frac{Weight}{Molar Mass \times Volume}\), find the concentration of the malonic acid. The molar mass is derived from Step 3, molarity of malonic acid M1 = \( \frac{19.5g}{104g/mol \times 0.250L} \)

Step 5: Determine the Acid Ionization Constants \(K_{a_{1}}\)

Use the formula for finding the acid ionization constant \(K_{a_{1}}\). \( K_{a_{1}} = \frac{[H^+][Hmal^-]}{[Hmal]} = \frac{[H^+]^2}{M1 - [H^+]}\) Where: [Hmal^-] and [H^+] are approximately equal as for every mole of malonic acid that ionizes, one mole of H ions and one mole of Hmal- ions are produced. Thus their concentrations are equal. Furthermore, [Hmal] = M1 - [H^+] because the initial concentration of the malonic acid is M1 and as it ionizes, its concentration decreases by the same amount as [H^+].

Step 6: Determine the Acid Ionization Constants \(K_{a_{2}}\)

From Step 2, the [H+] and hence [Mal^2-] for the second ionization may be calculated for the sodium hydrogen malonate solution. Therefore, using the equation \( K_{a_{2}} = \frac{[H^+][Mal^{2-}]}{[Hmal^-]} = \frac{[H^+]^2}{M2 - [H^+]}\). Therefore, we can calculate \(K_{a_{2}}\) by substituting the known values.

Key Concepts

Malonic Acid

Malonic acid, also known as propanedioic acid, is an organic compound with the formula \( \text{C}_3\text{H}_4\text{O}_4 \). It plays a significant role in chemistry due to its ability to act as a dicarboxylic acid, meaning it contains two carboxyl (-COOH) groups.
Malonic acid is particularly interesting for its ability to undergo partial ionization in water, leading to the release of protons (\( \text{H}^+ \) ions). This property is essential for determining the acid ionization constants, \( K_{a_{1}} \) and \( K_{a_{2}} \), which indicate the strength of the acid's dissociation at different stages.

• The first ionization removes a proton from one of the carboxyl groups, forming hydrogen malonate ions (\( \text{HMal}^- \)).
• The second ionization, which occurs to a lesser extent, removes a proton from the second carboxyl group, forming malonate ions (\( \text{Mal}^{2-} \)).

Understanding these ionization processes is crucial for calculations involving acid strength and equilibrium in aqueous solutions.

pH Calculation

pH is a measure of the acidity of a solution, representing the concentration of hydrogen ions (\( [\text{H}^+] \)). It is calculated using the formula: \[ pH = -\log[\text{H}^+] \] This formula shows that the smaller the pH value, the higher the concentration of \( [\text{H}^+] \) ions, indicating a stronger acidic nature.
For instance, in the analysis of malonic acid, two distinct pH values were given: 1.47 for the malonic acid solution and 4.26 for sodium hydrogen malonate solution. These values are crucial in finding the ionization constants:

• The pH of 1.47 indicates a higher concentration of \( [\text{H}^+] \), showing significant ionization of malonic acid in water.
• A pH of 4.26 suggests a lower \( [\text{H}^+] \) concentration, corresponding to the weaker ionization of sodium hydrogen malonate.

pH calculation is thus pivotal in understanding how different compounds behave in solutions and how they affect equilibrium.

Sodium Hydrogen Malonate

Sodium hydrogen malonate is an important compound in studying the ionization of malonic acid because it acts as an intermediate in its ionization process. It consists of a sodium ion (\( \text{Na}^+ \)) and a hydrogen malonate ion (\( \text{HMal}^- \)).
In an aqueous solution, sodium hydrogen malonate further ionizes to form malonate ions (\( \text{Mal}^{2-} \)) and releases additional \( \text{H}^+ \) ions, contributing to the solution's acidity.

• This compound is crucial for assessing \( K_{a_{2}} \), the second acid ionization constant of malonic acid.
• By examining its pH, one can deduce the extent to which the second ionization has occurred, helping in the calculation of \( K_{a_{2}} \).

Understanding the role of sodium hydrogen malonate helps in mapping the complete ionization profile of malonic acid.

Ionization Equilibrium

Ionization equilibrium involves the balance that is established between the ionized and non-ionized forms of an acid in solution. This balance is significantly influenced by the strength of the acid and its acid ionization constants, \( K_{a_{1}} \) and \( K_{a_{2}} \).
Malonic acid has two ionization steps because it possesses two acidic protons. The equilibrium expression for these steps are as follows:

• For the first ionization: \[K_{a_{1}} = \frac{[\text{H}^+][\text{HMal}^-]}{[\text{H}_2\text{Mal}]} \] This represents the equilibrium between malonic acid and its first ionized form, \( \text{HMal}^- \).
• For the second ionization:\[ K_{a_{2}} = \frac{[\text{H}^+][\text{Mal}^{2-}]}{[\text{HMal}^-]} \]This equation describes the equilibrium between the first and second ionized forms of the acid.

Ionization equilibrium is key in predicting the behavior of acids in various conditions, including their reactivity and interactions with other substances in a solution.