Chapter 16: Problem 55
A particular sample of vinegar has a pH of 2.20 . 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 \(6.10 \times 10^{-3} \mathrm{M}\).
Step by step solution
01
Write down the pH equation and CH3COOH dissociation equation
The pH equation is:
\(pH = -\log_{10} [\mathrm{H}^{+}]\)
The dissociation equation for acetic acid is:
\(\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^{-}} + \mathrm{H^+}\)
02
Calculate the concentration of H+ ions
We are given the pH of the solution, which is 2.20. We can use the pH equation to find the concentration of H+ ions:
\(2.20 = -\log_{10} [\mathrm{H}^{+}]\)
Rearrange the equation to solve for the concentration of H+ ions:
\([\mathrm{H}^{+}] = 10^{-2.20}\)
03
Calculate the concentration of acetic acid
Now, we need to use the dissociation equation for acetic acid and the given Ka value to calculate the concentration of acetic acid. Since the acetic acid is the only acid present in the vinegar, we can assume that the concentration of H+ ions and CH3COO- ions are equal.
Let's write the Ka expression for acetic acid dissociation:
\(K_a = \dfrac{[\mathrm{CH_3COO^{-}}][\mathrm{H}^{+}]}{[\mathrm{CH_3COOH}]}\)
Given that \([\mathrm{CH_3COO^{-}}] = [\mathrm{H}^{+}]\), the Ka expression becomes:
\(K_a = \dfrac{[\mathrm{H}^{+}]^2}{[\mathrm{CH_3COOH}]}\)
We already calculated the concentration of H+ ions in step 2. Now we can solve for the concentration of acetic acid.
Rearrange the equation:
\([\mathrm{CH_3COOH}] = \dfrac{[\mathrm{H}^{+}]^2}{K_a}\)
Substitute the known values:
\([\mathrm{CH_3COOH}] = \dfrac{(10^{-2.20})^2}{1.8 \times 10^{-5}}\)
04
Compute the concentration of acetic acid
Now we can compute the concentration of acetic acid using the obtained values:
\([\mathrm{CH_3COOH}] = \dfrac{(10^{-2.20})^2}{1.8 \times 10^{-5}} \approx 6.10 \times 10^{-3} \mathrm{M}\)
So, the concentration of acetic acid in the vinegar is approximately \(6.10 \times 10^{-3} \mathrm{M}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
pH Calculation
Understanding pH is crucial when working with acids like acetic acid found in vinegar. The pH scale, ranging from 0 to 14, measures how acidic or basic a solution is. A pH of 7 is neutral, below 7 is acidic, and above 7 is basic. To calculate the pH of a solution, we use the formula:
In this exercise, the vinegar has a pH of 2.20. By rearranging the equation, we can find the hydrogen ion concentration:
It's important to remember that each decrease in pH by one unit represents a tenfold increase in hydrogen ion concentration, highlighting how potent even small changes in pH can be.
- \( pH = -\log_{10} [\mathrm{H}^{+}] \)
In this exercise, the vinegar has a pH of 2.20. By rearranging the equation, we can find the hydrogen ion concentration:
- \([\mathrm{H}^{+}] = 10^{-2.20}\)
It's important to remember that each decrease in pH by one unit represents a tenfold increase in hydrogen ion concentration, highlighting how potent even small changes in pH can be.
Acid Dissociation Constant
The acid dissociation constant, \(K_a\), is a specific constant that indicates the strength of an acid in a solution. For acetic acid, \(K_a\) is given as \(1.8 \times 10^{-5}\). This means acetic acid is a weak acid, as strong acids have larger \(K_a\) values.
Understanding \(K_a\) helps us know how much the acid dissociates into ions in a solution. For acetic acid, the dissociation can be represented as:
By solving this equation, we understand the extent of acetic acid dissociation, which is essential for calculating acetic acid concentration in solutions like vinegar.
Understanding \(K_a\) helps us know how much the acid dissociates into ions in a solution. For acetic acid, the dissociation can be represented as:
- \( \mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^{-}} + \mathrm{H^+} \)
- \( K_a = \dfrac{[\mathrm{CH_3COO^{-}}][\mathrm{H}^{+}]}{[\mathrm{CH_3COOH}]} \)
By solving this equation, we understand the extent of acetic acid dissociation, which is essential for calculating acetic acid concentration in solutions like vinegar.
Vinegar Composition
Vinegar is a common household item primarily composed of water and acetic acid. This acetic acid gives vinegar its sour taste and distinctive smell. The concentration of acetic acid in vinegar usually ranges from about 4% to 8% by volume.
In chemical terms, vinegar is an aqueous solution of acetic acid. For solutions where acetic acid is the only significant acid, its concentration can be calculated using the principles of pH and \(K_a\).
The exercise above illustrated how to determine the concentration of acetic acid when given pH and \(K_a\). By finding the hydrogen ion concentration through pH, we can calculate the acetic acid concentration with:
In chemical terms, vinegar is an aqueous solution of acetic acid. For solutions where acetic acid is the only significant acid, its concentration can be calculated using the principles of pH and \(K_a\).
The exercise above illustrated how to determine the concentration of acetic acid when given pH and \(K_a\). By finding the hydrogen ion concentration through pH, we can calculate the acetic acid concentration with:
- \( [\mathrm{CH_3COOH}] = \dfrac{[\mathrm{H}^{+}]^2}{K_a} \)