Question

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), received is name because it is present in sour milk as a product of bacterial action. It is also responsible for the soreness in muscles after vigorous exercise. (a) The \(\mathrm{pK}_{a}\) of lactic acid is \(3.85\). Compare this with the value for propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}, \mathrm{p} K_{a}=4.89\right)\), and explain the differ- ence. (b) Calculate the lactate ion concentration in a \(0.050 \mathrm{M}\) solution of lactic acid. (c) When a solution of sodium lactate, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\), is mixed with an aqueous copper(II) solution, it is possible to obtain a solid salt of copper(II) lactate as a blue-green hydrate, \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COO}\right)_{2} \mathrm{Cu} \cdot x \mathrm{H}_{2} \mathrm{O} .\) Elemental analysis of the solid tells us that the solid is \(22.9 \% \mathrm{Cu}\) and \(26.0 \% \mathrm{C}\) by mass. What is the value for \(x\) in the formula for the hydrate? (d) The acid-dissociation constant for the \(\mathrm{Cu}^{2+}(a q)\) ion is \(1.0 \times 10^{-8} .\) Based on this value and the acid-dissociation constant of lactic acid, predict whether a solution of copper(II) lactate will be acidic, basic, or neutral. Explain your answer.

Short answer

The difference in pKa values between Lactic acid and Propionic acid is due to the presence of an OH group in lactic acid, which strengthens its acidity. The lactate ion concentration in a 0.050 M lactic acid solution can be calculated using the Ka value and equilibrium constant equation. Using the given percentages of Cu and C in the copper(II) lactate hydrate, we can determine the molar mass and the value of x (moles of water) in its formula. Since the acid-dissociation constant of lactic acid is larger than that of Cu²⁺, a solution of copper(II) lactate will be acidic.

Step-by-step solution

(a) Comparing pKa values and explaining the difference

The given pKa values are: pKa of Lactic acid \((\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}) = 3.85\) and pKa of Propionic acid \((\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}) = 4.89\). The lower the value of pKa, the stronger the acid. Hence lactic acid is a stronger acid as compared to propionic acid. The difference lies in the presence of \(\mathrm{OH}\) group on the \(\mathrm{CH}\) of lactic acid, which strengthens the acidity by allowing more charge delocalization due to intramolecular hydrogen bonding when the acid dissociates.

(b) Calculating lactate ion concentration

We have a \(0.050 \mathrm{M}\) solution of lactic acid. Using the pKa value, we can calculate the \(\mathrm{Ka}\) value as follows: \(\mathrm{Ka} = 10^{-\mathrm{pKa}} = 10^{-3.85}\) Now, let the lactate ion concentration be denoted by x. Then we can write the equation for the equilibrium constant: \(\mathrm{Ka} = \frac{x^{2}}{0.050 - x}\) Solving for x, we get the lactate ion concentration.

(c) Finding x in the formula for the hydrate

The percentage of Cu and C in the hydrate are given as 22.9% and 26.0% respectively. Let the molar mass of the whole compound be M and the value of x be the moles of water present in the hydrate. Then, we can calculate M using the given percentages of Cu and C in the compound: \(M = \frac{63.55}{0.229} + \frac{12}{0.26}\) Now, let's calculate the molar mass of the anhydrous part (without water) of the copper(II) lactate: \((\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COO})_{2} \mathrm{Cu}\), its molar mass is \(2 \times[\mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{COO}]+[\mathrm{Cu}]\). Using the total molar mass (M) and the molar mass of anhydrous part, we can find the value of x in the formula for the hydrate.

(d) Predicting the acidity of the solution

We are given that the acid-dissociation constant for \(\mathrm{Cu}^{2+}(a q)\) ion is \(1.0 \times 10^{-8}\), and the acid-dissociation constant of lactic acid is \(10^{-3.85}\). Since the acid-dissociation constant of lactic acid is larger than that of \(\mathrm{Cu}^{2+}\), lactic acid is the stronger acid. So, when both are present in the same solution, there is only partial dissociation of \(\mathrm{Cu}^{2+}\) leaving the solution primarily acidic. Hence, we predict that a solution of copper(II) lactate will be acidic.

Key concepts

Understanding Lactic Acid

Lactic acid, often encountered in daily life from sour milk to muscle soreness, is an organic compound with the chemical formula \(\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}\). To truly appreciate its role in chemistry and biology, one must first understand its acidic properties.

Lactic acid's \(\mathrm{pK}_a\) value, which stands at 3.85, is imperative to note. This value indicates it's a relatively strong acid, especially when compared to propionic acid with a \(\mathrm{pK}_a\) of 4.89. This difference in acidity can be attributed to the \(\mathrm{OH}\) group in lactic acid, which through intramolecular hydrogen bonding, enhances the ability of the molecule to stabilize the negative charge formed upon losing a proton. This stabilization increases the tendency of lactic acid to donate a proton, thus increasing its acidity.

Understanding lactic acid's acidic nature not only helps in conducting equilibrium constant calculations but also lays the groundwork for comprehending its salts, like copper(II) lactate, and their behaviors in solutions.

Equilibrium Constant Calculation for Acid-Base Reactions

The equilibrium constant, denoted as \(K_a\) for acids, is a quantitative measure of their strength – the tendency of an acid to donate a proton to its conjugate base. Calculating \(K_a\) from the given \(\mathrm{pK}_a\) is a crucial step in determining the concentrations of species in an acid-base equilibrium.

To embark on this journey, let's ascertain the \(K_a\) of lactic acid using the formula \(K_a = 10^{-\mathrm{pK}_a}\). Once calculated, we utilize an expression that aligns with the acid dissociation equilibrium: \(K_a = \frac{x^2}{0.050 - x}\), where x represents the lactate ion concentration. Solving this equation allows us to compute the concentration of lactate ions in solution, which is central to understanding the behavior of lactic acid in biological systems and industrial applications.

Through this method, we navigate the complexity of equilibrium principles, empowering students to approach similar problems with confidence.

Characteristics and Analysis of Copper(II) Lactate

Diving into the world of coordination compounds, copper(II) lactate emerges when sodium lactate interacts with an aqueous copper(II) solution. The resulting blue-green hydrate, intriguing in its properties, presents an excellent case study for stoichiometry and chemical analysis.

Elemental analysis provides a pathway to unlocking the formula of this fascinating compound. By applying the given mass percentages of copper and carbon, we can derive the empirical formula of the hydrate. This empirical deduction entails calculating the formula mass using copper's and carbon's contributions, then deducing the number of water molecules, symbolized by x, in the hydrate's structure.

Understanding compounds like copper(II) lactate is not restricted to their formulation. Predicting the pH of their solutions involves juxtaposing their \(K_a\) with other species in the mixture — an endeavor that encapsulates the principles of acid-base chemistry and solubility equilibrium to determine whether the resultant solution is acidic, neutral, or basic.