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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) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}{ }^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}\).

(a) What is a strong acid? (b) A solution is labeled \(0.500 \mathrm{M} \mathrm{HCl}\). What is \(\left[\mathrm{H}^{+}\right]\) for the solution? (c) Which of the following are strong acids: \(\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}, \mathrm{HI}\) ?

(a) Which of the following is the stronger BronstedLowry acid, \(\mathrm{HNO}_{3}\) or \(\mathrm{HNO}_{2} ?\) (b) Which is the stronger Br酶nsted- Lowry base, \(\mathrm{NH}_{3}\) or \(\mathrm{H}_{2} \mathrm{O}\) ? Briefly explain your choices.

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(\mathrm{aq})\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l)\). (c) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{7} ?\) What is its conjugate base?

A \(0.100 \mathrm{M}\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is \(11.0 \%\) ionized. Using this information, calculate \(\left.\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right)\right]\), and \(K_{a}\) for chloroacetic acid.
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