Problem 164 - Chemistry An Atoms First Approach

Question

Consider a \(0.60-M\) solution of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3},\) lactic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.4 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) ii. \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) iii. \(\mathrm{H}^{+}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) \(\mathbf{v} . \mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x,\) the amount (mol/L) of lactic acid that dissociates to reach equilibrium. c. What is the equilibrium concentration for \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-} ?\) d. Calculate the \(p H\) of the solution.

Step-by-Step Solution

Answer

The major species present in the solution are HC₃H₅O₃, C₃H₅O₃⁻, H⁺, and H₂O. The equilibrium concentration of C₃H₅O₃⁻ is approximately 0.0093 M. The pH of the 0.60 M lactic acid solution is approximately 2.03.

Step 1 - Write down the dissociation equilibrium

Lactic acid, HC₃H₅O₃,(a weak acid), will dissociate into the hydrogen ion, H⁺, and the lactate ion, C₃H₅O₃⁻. Write down the dissociation equilibrium as follows: \[HC_{3}H_{5}O_{3} \rightleftharpoons H^{+} + C_{3}H_{5}O_{3}^{-}\]

Step 2 - Identify the major species in the solution

All species involved in dissociation of the lactic acid as well as water are considered major species in the solution. Thus, the major species present in the solution are i. HC₃H₅O₃ ii. C₃H₅O₃⁻ iii. H⁺ iv. H₂O

Step 3 - Set up the ICE table

Create an ICE table to track initial, change, and equilibrium concentrations of all species: | | HC₃H₅O₃ | H⁺ | C₃H₅O₃⁻ | |---------------|-------------|------------|-----------| | Initial (M) | 0.60 | 0 | 0 | | Change (M) | -x | +x | +x | | Equilibrium(M)| 0.60 - x | x | x |

Step 4 - Write the Ka expression

Use the Ka = 1.4 × 10⁻⁴ and the equilibrium concentrations from the ICE table to write the Ka expression: \[1.4 \times 10^{-4} = \frac{x \times x}{0.60 - x}\]

Step 5 - Solve for x

Since Ka is small, we can approximate that x << 0.60, so 0.60 - x is nearly equal to 0.60. Hence, we can simplify and solve for x as follows: \[1.4 \times 10^{-4} = \frac{x^2}{0.60}\] \[x^2 = 1.4 \times 10^{-4} \times 0.60\] \[x = \sqrt{1.4 \times 10^{-4} \times 0.60}\] \[x \approx 0.0093 \,\text{M}\]

Step 6 - Find the equilibrium concentration of C₃H₅O₃⁻

The equilibrium concentration of C₃H₅O₃⁻, from our ICE table, is equal to x: \[ [C_{3}H_{5}O_{3}^{-}]_{eq} = x \approx 0.0093\,\text{M}\]

Step 7 - Calculate the pH of the solution

Since x is also the concentration of H⁺, calculate the pH using the formula pH = -log[H⁺]: \[pH = -\log(0.0093) \approx 2.03\] So, the pH of the 0.60 M lactic acid solution is approximately 2.03.

Key Concepts

Acid Dissociation Constant (Ka)

Understanding the acid dissociation constant, often abbreviated as Ka, is crucial when studying chemistry, particularly acid-base reactions. Ka measures the strength of an acid in solution, representing the equilibrium constant for the dissociation of an acid into its conjugate base and a proton (H+). A higher Ka value denotes a stronger acid, meaning that the acid donates protons more readily.

For lactic acid (HC₃H₅O₃), the dissociation can be represented by the equation:
\[HC_{3}H_{5}O_{3} \rightleftharpoons H^{+} + C_{3}H_{5}O_{3}^{-}\]
Here, the Ka value is given as \(1.4 \times 10^{-4}\), which suggests that lactic acid is a weak acid, since its Ka value is relatively low, indicating it only partially dissociates in solution. When comparing to strong acids—with Ka values much higher—lactic acid produces fewer protons and its conjugate base in the solution at equilibrium.

ICE Table

The ICE table is an invaluable tool used to track the changes in concentrations of species in a chemical reaction, from the Initial state, through the Change, to the Equilibrium state. To illustrate this, let's use the dissociation of lactic acid.

Initially, we have a 0.60-M solution of lactic acid with no significant amount of H+ or lactate ions (C₃H₅O₃⁻). Upon dissociation, a certain amount \(x\) of lactic acid is converted into \(x\) moles per liter of H+ and \(x\) moles per liter of lactate ions. The ICE table helps us organize these thoughts:

• Initial: Lactic acid at 0.60 M, H+ and lactate ions both at 0 M
• Change: Lactic acid decreases by \(x\) M, H+ and lactate ions increase by \(x\) M
• Equilibrium: Lactic acid at \(0.60 - x\) M, H+ and lactate ions both at \(x\) M

By setting up this framework, we can then apply our understanding of the acid dissociation constant to find the value of \(x\), which indicates how much lactic acid has dissociated at equilibrium.

pH Calculation

Calculating the pH of a solution is essential for understanding its acidity. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:\[pH = -\log[H^+]\]. In our example, \(x\) represents the equilibrium concentration of hydrogen ions produced by the dissociation of lactic acid.

To calculate the pH, we use the concentration of hydrogen ions that we obtained from solving the ICE table and insert it into the pH equation. For lactic acid, after approximating and simplifying the Ka expression, we found that \(x\) was approximately 0.0093 M, which is the concentration of hydrogen ions at equilibrium. Thus, the pH of the solution can be calculated:
\[pH = -\log(0.0093) \approx 2.03\]
As reflected by a pH of 2.03, the solution is acidic. This value is significant as it correlates with the weak acidic nature of lactic acid, which only partially dissociates in solution, releasing enough H+ to lower the pH below the neutral value of 7.