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Chapter 16: Problem 55

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. 16.56 If a solution of \(\mathrm{HF}\left(K_{a}=6.8 \times 10^{-4}\right)\) has a \(\mathrm{pH}\) of \(3.65,\) calculate the concentration of hydrofluoric acid.

Short Answer

Expert verified
The concentration of acetic acid in the vinegar is approximately \(8.96 \times 10^{-3} \, \text{M}\).

Step by step solution

01

Calculate the concentration of H+ ions using the pH formula

We know that the formula for pH is given by: \(pH = -\log{[H^{+}]}\) where [H+] is the concentration of H+ ions. In this problem, we are given the pH as 2.90. We can rewrite the formula to find the concentration of H+ ions: \([H^{+}] = 10^{-pH}\) Now, plugging in the given pH value: \([H^{+}]= 10^{-2.90}\)
02

Calculate the concentration of H+ ions

By evaluating the expression above, we get: \([H^{+}] = 1.26 \times 10^{-3} \, \text{M}\)
03

Write the Ka expression for acetic acid

The equilibrium expression for acetic acid (CH3COOH) reacting with water can be written as: \(K_a = \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\) Given that the concentration of H+ ions is small compared to the acetic acid concentration, we can approximate that: \([CH_{3}COO^{-}] \approx [H^{+}]\) and \([CH_{3}COOH] \approx C - [H^{+}]\) where C is the initial concentration of acetic acid. Now, substituting these values into the Ka expression: \(1.8 \times 10^{-5} = \frac{(1.26 \times 10^{-3})(1.26 \times 10^{-3})}{(C - 1.26 \times 10^{-3})}\)
04

Calculate the concentration of acetic acid

Now, we just need to solve for the concentration of acetic acid (C): \(C - 1.26 \times 10^{-3} = \frac{(1.26 \times 10^{-3})^2}{1.8 \times 10^{-5}}\) After solving for C, we get: \(C = 8.96 \times 10^{-3} \, \text{M}\) Thus, the concentration of acetic acid in the vinegar is approximately \(8.96 \times 10^{-3} \, \text{M}\).

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Most popular questions from this chapter

A \(0.100 M\) solution of bromoacetic acid \(\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\) is \(13.2 \%\) ionized. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\) and \(K_{a}\) for bromoacetic acid.

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (a) What A \(0.035 \mathrm{M}\) solution of ephedrine has a pH of 11.33 . are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}\), \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\) and \(\mathrm{OH}^{-} ?\) (b) Calculate \(K_{b}\) for ephedrine.

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-},\) (b) \(\mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-},(\mathbf{c})\) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.724 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right) .\) What is the \(\mathrm{pH}\) of this solution?

(a) Which of the following is the stronger Bronsted-Lowry acid, HBrO or HBr? (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{F}^{-}\) or \(\mathrm{Cl}^{-}\) ? Briefly explain your choices.
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