A typical sample of vinegar has a pH of \(3.0\). Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

Short Answer

Expert verified
The concentration of acetic acid in the vinegar is approximately 0.056 M.

Step by step solution

01

Write down the given values

We are given: - pH of vinegar: 3.0 - Ka of acetic acid: \(1.8 \times 10^{-5}\)
02

Calculate the hydrogen ion concentration

We will first calculate the hydrogen ion concentration by using the pH formula. The formula to find the hydrogen ion concentration given the pH value is: \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} \] So by plugging in our pH value, we get: \[ [\mathrm{H^+}] = 10^{-3} \] \[ [\mathrm{H^+}] = 1 \times 10^{-3}\ \text{M} \]
03

Write the Ka expression

Now we need to write the Ka expression for acetic acid (CH3COOH). The general form of the Ka expression is: \[ K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\] Where \([\mathrm{H^+}]\) is the concentration of hydrogen ions, \([\mathrm{A^-}]\) is the concentration of acetate ions, and \([\mathrm{HA}]\) is the concentration of acetic acid. In our case: \[1.8 \times 10^{-5} = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}\]
04

Use initial and equilibrium concentrations

Let the initial concentration of acetic acid be C (in moles per liter). Since all dissociation should come from the acetic acid, we can write the concentrations at equilibrium as follows: - [CH3COOH] = C - \( [\mathrm{H^+}] \) - [CH3COO-] = \( [\mathrm{H^+}] \) Now, plug these expressions into the Ka equation: \[1.8 \times 10^{-5} = \frac{([\mathrm{H^+}])([\mathrm{H^+}])}{(C - [\mathrm{H^+}])}\]
05

Solve for the concentration of acetic acid

Now, we have the following equation: \[1.8 \times 10^{-5} = \frac{(1 \times 10^{-3})(1 \times 10^{-3})}{(C - 1 \times 10^{-3})}\] Now, let's solve for C: \[ 1.8 \times 10^{-5} = \frac{1 \times 10^{-6}}{(C - 1 \times 10^{-3})}\] \[ C - 1 \times 10^{-3} = \frac{1 \times 10^{-6}}{1.8 \times 10^{-5}}\] \[ C = 1 \times 10^{-3} + \frac{1 \times 10^{-6}}{1.8 \times 10^{-5}}\] \[ C = 1 \times 10^{-3} + 5.56 \times 10^{-2}\] \[ C = 0.056\ \text{M} \] The concentration of acetic acid in the vinegar is approximately 0.056 M.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

pH of Vinegar
Understanding the pH of vinegar is crucial when we study its acidic properties. The pH scale is used to determine the acidity or basicity of an aqueous solution and ranges from 0 to 14. A pH of 7 is considered neutral, while a pH less than 7 indicates an acidic solution, and a pH greater than 7 indicates a basic solution.

Vinegar, which is primarily a dilute solution of acetic acid in water, typically has a pH around 3.0. This means vinegar is quite acidic. The pH value is used to calculate the hydrogen ion concentration \( [\mathrm{H^+}] \) in the solution, which is a direct measure of its acidity. The concentration of hydrogen ions is inversely related to the pH value and can be calculated using the equation \( [\mathrm{H^+}] = 10^{-\mathrm{pH}} \).

In the context of acetic acid in vinegar, knowing the pH allows us to deduce how much the acid has dissociated in the solution. It is this dissociation that gives vinegar its acidic properties and determines the concentration of acetic acid in the solution.
Ka Expression for Acetic Acid
Acetic acid, with the chemical formula \(\mathrm{CH_3COOH}\), is a weak acid that partially dissociates in water to form hydrogen ions \(\mathrm{H^+}\) and acetate ions \(\mathrm{CH_3COO^-}\). The degree to which acetic acid dissociates is quantified by its acid dissociation constant, \(K_a\). The \(K_a\) expression is written as:

\[ K_a = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} \]
This equilibrium expression relates the concentration of hydrogen ions and acetate ions to the concentration of undissociated acetic acid at equilibrium. It provides a numerical value that helps determine how strong or weak an acid is; the smaller the \(K_a\), the weaker the acid.

For acetic acid, the \(K_a\) value is \(1.8 \times 10^{-5}\), reflecting its weak acid nature. This information is essential when solving for the equilibrium concentration of acetic acid in a vinegar solution.
Equilibrium Concentration of Acetic Acid
Determining the equilibrium concentration of acetic acid in vinegar involves using the \(K_a\) expression mentioned earlier. The equilibrium concentration tells us how much acetic acid remains undissociated at equilibrium.

To solve for the equilibrium concentration, we can assume an initial concentration of acetic acid, \(C\), and express the concentrations of hydrogen and acetate ions in terms of \(C\) and the known \(\mathrm{H^+}\) concentration from the pH of the solution. Given that the concentration of the acetate ion \(\mathrm{CH_3COO^-}\) is equal to the concentration of \(\mathrm{H^+}\) and the concentration of acetic acid \(\mathrm{CH_3COOH}\) is the initial concentration minus the \(\mathrm{H^+}\) concentration, you can set up the equation:

\[ K_a = \frac{(\mathrm{H^+})(\mathrm{H^+})}{(C - \mathrm{H^+})} \]
This equation will then be rearranged and solved for \(C\), giving the equilibrium concentration of acetic acid in the vinegar solution. This concentration is key to understanding the potency of vinegar and has practical implications in cooking, cleaning and other industrial processes where vinegar's acidic properties are utilized.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are the major species present in a \(0.150-M \mathrm{NH}_{3}\) solution? Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of this solution.

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. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x\), the amount \((\mathrm{mol} / \mathrm{L})\) 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 \(\mathrm{pH}\) of the solution.

Quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{2}\right)\) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, \(\mathrm{p} K_{\mathrm{b}_{\mathrm{b}}}=5.1\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .\) Only \(1 \mathrm{~g}\) quinine will dissolve in \(1900.0 \mathrm{~mL}\) of solution. Calculate the \(\mathrm{pH}\) of a saturated aqueous solution of quinine. Consider only the reaction \(\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}\) described by \(\mathrm{p} K_{\mathrm{b}_{1}}\), where \(\mathrm{Q}=\) quinine.

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3}\). The \(K_{\mathrm{a}}\) value for hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

A typical aspirin tablet contains \(325 \mathrm{mg}\) acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right) .\) Calculate the \(\mathrm{pH}\) of a solution that is prepared by dissolving two aspirin tablets in enough water to make one cup \((237 \mathrm{~mL})\) of solution. Assume the aspirin tablets are pure acetylsalicylic acid, \(K_{\mathrm{a}}=3.3 \times 10^{-4}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free