For a solution of \(0.025 M\) lactic acid, \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) \(\left(K_{\mathrm{a}}=8.4 \times 10^{-4}\right)\), calculate: (a) \(\left[\mathrm{H}^{+}\right]\) (b) \(\mathrm{pH}\) (c) percent ionization

Short Answer

Expert verified
[H^+] = 0.00458 M, pH = 2.34, Percent ionization = 18.32%

Step by step solution

01

- Write the ionization equation

Lactic acid, \(\text{HC}_3\text{H}_5\text{O}_2\), ionizes in water according to the equation: \[ \text{HC}_3\text{H}_5\text{O}_2 (aq) \leftrightharpoons \text{H}^+ (aq) + \text{C}_3\text{H}_5\text{O}_2^- (aq) \]
02

- Set up the ICE table

Initial, Change, Equilibrium (ICE) table: \[ \begin{array}{|c|c|c|c|} \hline & \text{HC}_3\text{H}_5\text{O}_2 & \text{H}^+ & \text{C}_3\text{H}_5\text{O}_2^- \ \hline \text{Initial (M)} & 0.025 & 0 & 0 \ \hline \text{Change (M)} & -x & +x & +x \ \hline \text{Equilibrium (M)} & 0.025 - x & x & x \ \hline \end{array} \]
03

- Apply the acid dissociation constant expression

The expression for the acid dissociation constant \(K_a\) is: \[ K_a = \frac{[\text{H}^+][\text{C}_3\text{H}_5\text{O}_2^-]}{[\text{HC}_3\text{H}_5\text{O}_2]} = \frac{x \cdot x}{0.025 - x} = 8.4 \times 10^{-4} \]
04

- Approximate if possible and solve for x

Since \( K_a \) is small, the approximation \( 0.025 - x \approx 0.025 \) can be made: \[ 8.4 \times 10^{-4} = \frac{x^2}{0.025} \] Solving for x: \(\begin{align*} x^2 &= (8.4 \times 10^{-4}) \cdot 0.025 \ x^2 &= 2.1 \times 10^{-5} \ x &= \sqrt{2.1 \times 10^{-5}} \ x &= 0.00458 \end{align*} \) Thus, \( [\text{H}^+] = 0.00458 \text{M} \)
05

- Calculate the pH

The pH is found by taking the negative logarithm (base 10) of the \( [\text{H}^+] \): \[ \text{pH} = -\log(0.00458) \approx 2.34 \]
06

- Calculate the percent ionization

Percent ionization is calculated by the formula: \[ \text{Percent Ionization} = \left( \frac{[\text{H}^+]}{\text{Initial Concentration}} \right) \times 100 = \left( \frac{0.00458}{0.025} \right) \times 100 \approx 18.32 \% \]

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Key Concepts

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

ICE table
An ICE table is a useful tool to keep track of the changes in concentrations of species involved in a chemical equilibrium. ICE stands for Initial, Change, and Equilibrium.
Here’s how it works:
  • Initial: Start with the initial concentrations of all reactants and products. For lactic acid ionization, initially, we have 0.025 M of lactic acid and 0 M of H+ and C3H5O2-.
  • Change: Determine the change that occurs as the system moves toward equilibrium. Represent these changes with a variable, often x. For lactic acid, it dissociates into H+ and C3H5O2-, so the change is -x for lactic acid and +x for both ions.
  • Equilibrium: Express the equilibrium concentrations in terms of the initial concentrations and the changes. For lactic acid, the equilibrium concentration is 0.025 - x, and for the ions, it’s x.
Using an ICE table simplifies solving equilibrium problems, especially involving weak acids or bases.
acid dissociation constant
The acid dissociation constant, denoted as Ka, measures the strength of a weak acid in solution. It represents the equilibrium constant for the ionization of the acid: \[ \text{HC}_3\text{H}_5\text{O}_2 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{C}_3\text{H}_5\text{O}_2^- (aq) \]In this case,
\[K_a = \frac{[\text{H}^+][\text{C}_3\text{H}_5\text{O}_2^-]}{[\text{HC}_3\text{H}_5\text{O}_2]}\]
For lactic acid, Ka is given as 8.4 × 10−4. A small Ka value indicates a weaker acid and lesser degree of ionization.
The equilibrium concentrations derived from the ICE table are used to set up the Ka expression. This approach helps to solve for the concentration of H+, which in turn is useful for calculating pH and percent ionization.
percent ionization
Percent ionization tells us how much of the acid has ionized in solution. It is calculated using the formula:
\[ \text{Percent Ionization} = \frac{[\text{H}^+]}{\text{Initial Concentration}} \times 100 \]For lactic acid with an initial concentration of 0.025 M and found [H+] as 0.00458 M:
\[ \text{Percent Ionization} = \frac{0.00458}{0.025} \times 100 \ \text{Percent Ionization} \text{ is approximately } 18.32\text{%} \]
This means that 18.32% of the lactic acid has ionized in the solution. Percent ionization is significant as it provides insight into the extent of ionization of a weak acid, which is crucial in understanding the acid’s behavior in different conditions.
pH calculation
The pH of a solution is a measure of its hydrogen ion concentration and is defined as:
\[ \text{pH} = -\text{log}[\text{H}^+] \]
For our lactic acid solution, where [H+] is 0.00458 M:
\[ \text{pH} = -\text{log}(0.00458) \ \text{pH} \text{ is approximately } 2.34 \]
The pH scale ranges from 0 to 14 with lower values being more acidic. Here, a pH of 2.34 indicates a moderately acidic solution. This step is essential for understanding the acidic or basic nature of the solution and is crucial for applications in chemistry and biology.
lactic acid
Lactic acid, with the chemical formula HC3H5O2, is a weak acid commonly found in various biological processes. It is produced in muscle cells during strenuous exercise when oxygen levels are low and is responsible for the feeling of muscle fatigue.
In an aqueous solution, lactic acid ionizes to form hydrogen ions (H+) and lactate ions (C3H5O2-).
Understanding its ionization helps in computing key properties like pH and percent ionization.
Moreover, it's widely used in food, pharmaceutical, and cosmetic industries due to its acidity and ability to promote preservation and flavor.

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Most popular questions from this chapter

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