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Chapter 16: Problem 57

A particular sample of vinegar has a pH of \(2.90\). If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.

Short Answer

Expert verified
The concentration of acetic acid in the vinegar is approximately 0.0882 M.

Step by step solution

01

Calculate the concentration of H+ ions

First, we must find the concentration of H+ ions in the solution. This can be done by using the pH formula: \[pH = -\log[H+]\] Where pH is the pH of the solution and \([H+]\) is the concentration of H+ ions in the solution. Since we are given the pH value as 2.90, we can rewrite the formula as follows: \[2.90 = -\log[H+]\]
02

Solve for the concentration of H+ ions

Now, we'll solve for \([H+]\) to find the concentration of H+ ions in the solution. To do this, simply isolate \([H+]\) by taking the inverse logarithm of both sides of the equation: \[[H+] = 10^{-pH}\] \[[H+] = 10^{-2.90}\] Calculate the concentration of H+ ions: \[[H+] \approx 1.26 \times 10^{-3} M\]
03

Write the Ka expression for acetic acid

Now that we have the concentration of H+ ions in the solution, we will utilize the \(K_a\) expression relating the concentrations of acetic acid, acetate ion, and H+ ion. The reaction of acetic acid (\(CH_3COOH\)) with water can be written as: \[CH_3COOH \rightleftharpoons H^+ + CH_3COO^-\] And the corresponding \(K_a\) expression for this reaction would be: \[K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\] Given \(K_a = 1.8 \times 10^{-5}\) for acetic acid, we can write the Ka expression as: \[1.8 \times 10^{-5} = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\]
04

Assume that [H+] is equal to [CH3COO-]

Since acetic acid (\(CH_3COOH\)) is the only acid present in the vinegar, we can assume that the concentration of H+ ions is equal to the concentration of acetate ions (\(CH_3COO^-\)) in the reaction: \[[H^+] = [CH_3COO^-]\]
05

Substitute the values into the Ka expression

Plug in the values for \([H+] \text{ , } [CH_3COO^-] \text{ and } K_a\) in the Ka expression: \[1.8 \times 10^{-5} = \frac{(1.26 \times 10^{-3})(1.26 \times 10^{-3})}{[CH_3COOH]}\]
06

Solve for the concentration of acetic acid

Now, we'll solve for the concentration of acetic acid, \([CH_3COOH]\): \[[CH_3COOH] = \frac{(1.26 \times 10^{-3})(1.26 \times 10^{-3})}{1.8 \times 10^{-5}}\] Calculate the concentration of acetic acid: \[[CH_3COOH] \approx 8.82 \times 10^{-2} M\] Hence, the concentration of acetic acid in the vinegar is approximately 0.0882 M.

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Most popular questions from this chapter

(a) Identify the Bronsted-Lowry acid and the Br酶nstedLowry base in the following reaction: (b) Identify the Lewis acid and the Lewis base in the reaction. [Sections \(16.2\) and \(16.11]\)

A hypothetical acid \(\mathrm{H}_{2} \mathrm{X}\) is both a strong acid and a diprotic acid. (a) Calculate the pH of a \(0.050 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{X}\), assuming that only one proton ionizes peracid molecule. (b) Calculate the \(\mathrm{pH}\) of the solution from part (a), now assuming that both protons of each acid molecule completely ionize. (c) In an experiment it is observed that the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{X}\) is \(1.27 .\) Comment on the relative acid strengths of \(\mathrm{H}_{2} \mathrm{X}\) and \(\mathrm{HX}^{-}\). (d) Would a solution of the salt \(\mathrm{NaH} \mathrm{X}\) be acidic, basic, or neutral? Explain.

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}=\) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{a}=4.3 \times 10^{-3}\) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \({ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{b}=6.0 \times 10^{-5}\) (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}\) What assumptions did you need to make? (b) What is the pH of a \(0.050 \mathrm{M}\) aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with pH 13? With pH 1?

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid. or a species with negligible acidity: (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), (b) \(\mathrm{HCO}_{3}^{-}\), (c) \(\mathrm{O}^{2-}\), (d) \(\mathrm{Cl}^{-}\) (e) \(\mathrm{NH}_{3} .\)

How does the acid strength of an oxyacid depend on (a) the electronegativity of the central atom; (b) the number of nonprotonated oxygen atoms in the molecule?
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