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Lactic acid \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\right)\), which occurs in sour milk and foods such as sauerkraut, is a weak monoprotic acid. The \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of lactic acid is \(2.43 .\) What are the values of \(K_{a}\) and \(\mathrm{p} K_{a}\) for lactic acid?

Short Answer

Expert verified
\(K_a \approx 1.38 \times 10^{-4}\), \(pK_a \approx 3.86\).

Step by step solution

01

Understand the Components

First, identify that you're dealing with lactic acid, a weak monoprotic acid (C3H6O3), with a given concentration of 0.10 M and a pH of 2.43. We're asked to find the acid dissociation constant \(K_a\) and its logarithmic form \(pK_a\).
02

Calculate the Hydrogen Ion Concentration

To find the hydrogen ion concentration \([H^+]\), use the pH formula: \[\text{pH} = -\log[H^+]\]Given pH is 2.43, substitute and solve for \([H^+]\): \[[H^+] = 10^{-2.43}\] Calculate \([H^+] \approx 3.72 \times 10^{-3} \text{ M}\).
03

Use the Equilibrium Expression for a Weak Acid

For a weak acid HA dissociating in water:\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]The equilibrium expression is:\[K_a = \frac{[H^+][A^-]}{[HA]}\]Assuming initial concentration of lactic acid \( [\text{HA}] = 0.10 \text{ M} \), as it's weak, ignore decrease due to dissociation so \([HA] \approx 0.10 \text{ M}\). Then \([A^-] = [H^+]\) at equilibrium.
04

Compute the Acid Dissociation Constant \(K_a\)

Substitute the values into the equation:\[K_a = \frac{(3.72 \times 10^{-3})(3.72 \times 10^{-3})}{0.10}\]Calculate \(K_a \approx 1.38 \times 10^{-4}\).
05

Calculate \(pK_a\) From \(K_a\)

The \(pK_a\) is the negative logarithm of \(K_a\):\[pK_a = -\log(K_a)\]Substitute \(K_a \approx 1.38 \times 10^{-4}\):\[pK_a \approx 3.86\]Thus, \(pK_a \approx 3.86\).

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Key Concepts

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

Weak Acids
Weak acids are fascinating because they do not completely dissociate in water. Unlike strong acids, which fully break down into ions, weak acids only partially separate. This means that in a solution, most of the acid molecules remain intact, while only a small portion release hydrogen ions (\( H^+ \)). This characteristic behavior of weak acids can significantly affect their reactivity and the overall acidity of a solution. For our example, lactic acid, with the formula (\( ext{C}_3 ext{H}_6 ext{O}_3 \)), is a common weak acid found in sour milk and other fermented foods. When such an acid is in solution, it sets up an equilibrium between the undissociated acid molecules and the ions that it forms.Understanding how a weak acid behaves in a solution helps us predict the pH level and calculate the degree of dissociation just like we needed to do for lactic acid.
Equilibrium Expression
The equilibrium expression is a formula that describes the balance between reactants and products in a chemical reaction at equilibrium. For weak acids like lactic acid, this expression is crucial because it helps us understand how much of the acid has dissociated.In the dissociation of a weak acid, represented as \( ext{HA} ightleftharpoons ext{H}^+ + ext{A}^- \), the equilibrium constant, known as the acid dissociation constant (\( K_a \)), tells us the extent of the acid's ionization. The equilibrium expression is given by:\[K_a = \frac{[H^+][A^-]}{[HA]} \]This ratio involves the concentration of the hydrogen ions ([\( H^+ \)]), the conjugate base ions ([\( A^- \)]), and the undissociated acid ([\( HA \)]).In our scenario, for lactic acid, because it is weak, we assumed the initial concentration remained largely unchanged, making the calculation simpler. This expression is fundamental in determining the strength of the acid in equilibrium conditions.
pH Calculation
Calculating the pH of a solution is an essential skill in understanding acid-base chemistry, especially for weak acids. The pH scale runs from 0 to 14 and indicates how acidic or basic a solution is. A pH of 7 is neutral, lower values are acidic, and higher values are basic.To find the pH of a solution, the main formula used is:\[pH = -\log[H^+] \]For weak acids, calculating pH involves first determining the concentration of hydrogen ions (\( [H^+] \)) in a solution. Given a pH value, like 2.43 for a 0.10 M solution of lactic acid, we can work backwards to find \( [H^+] \):\[[H^+] = 10^{-2.43} \approx 3.72 \times 10^{-3} \text{ M} \]This hydrogen ion concentration is then used in equilibrium calculations to find the \( K_a \), and subsequently the \( pK_a \) of the acid, providing a complete picture of its dissociation behavior.These calculations show us how even a slight dissociation can affect the acid's pH and why understanding these steps is vital for analyzing weak acids.

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Most popular questions from this chapter

Choose from the conjugateacid-base pairs \(\mathrm{HSO}_{4}^{-} / \mathrm{SO}_{4}^{2-}, \mathrm{HF} / \mathrm{F}\). and \(\mathrm{NH}_{4}^{+} / \mathrm{NH}_{3}\), to complete the following equation with the pair that gives an equilibrium constant \(K_{c}>1\). $$ +\mathrm{NO}_{2}^{-} \longrightarrow \mathrm{HNO}_{2} $$

Use the conjugate acid-base pair \(\mathrm{HCN}\) and \(\mathrm{CN}^{-}\) to derive the relationship between \(K_{a}\) and \(K_{\mathrm{b}}\)

Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{HCO}_{3}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{H}_{3} \mathrm{O}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in \(0.010 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}\left(K_{\Delta 1}=4.3 \times 10^{-7} ; K_{\Delta 2}=5.6 \times 10^{-11}\right)\)

Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) in a solution prepared by mixing equal volumes of \(0.2 \mathrm{M} \mathrm{HCl}\) and \(0.6 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\left(K_{a 2}\right.\) for \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(\left.1.2 \times 10^{-2}\right)\)

Sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\right)\) is used as a food preservative. Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{Na}^{+}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, \mathrm{H}_{3} \mathrm{O}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in \(0.050 \mathrm{M}\) so- dium benzoate; \(K_{a}\) for benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right)\) is \(6.5 \times 10^{-5} .\)

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