Question

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

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

Step-by-step solution

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: \([H^{+}]= 10^{-2.90}\)

Calculate the concentration of H+ ions

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

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: \(1.8 \times 10^{-5} = \frac{(1.26 \times 10^{-3})(1.26 \times 10^{-3})}{(C - 1.26 \times 10^{-3})}\)

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 \(8.96 \times 10^{-3} \, \text{M}\).

Key concepts

pH Calculation

pH stands for the "potential of Hydrogen," and is a measure of how acidic or basic an aqueous solution is. It ranges from 0 to 14, with lower pH values indicating higher acidity, and higher pH values indicating basicity. Neutral pH is 7, which is pure water.
To calculate the pH of a solution, we use the formula:

• \[ pH = -\log{[H^+]} \]

In this formula, \([H^+]\) represents the concentration of hydrogen ions in moles per liter (M).
If the pH of the vinegar in our problem is 2.90, we can determine the concentration of hydrogen ions using this formula:

• \[ [H^+] = 10^{-pH} = 10^{-2.90} \]

By evaluating, you find the hydrogen ion concentration to be \(1.26 \times 10^{-3} \text{ M}\).
This straightforward calculation is crucial for understanding the acidity of a solution.

Acetic Acid Concentration

Acetic acid (\(\text{CH}_3\text{COOH}\)) is the primary component that gives vinegar its acidity. To find its concentration in the solution, we use its dissociation in water to form acetate ions (\(\text{CH}_3\text{COO}^-\)) and hydrogen ions (\(\text{H}^+\)):

• \( \text{CH}_3\text{COOH}_{(aq)} \leftrightarrow \text{CH}_3\text{COO}^-_{(aq)} + \text{H}^+_{(aq)} \)

Since we already know \([H^+]\) from the pH calculation, we can use the formula for the acid dissociation constant (\(K_a\)), tailored for acetic acid:

• \[ K_a = \frac{[\text{CH}_3\text{COO}^-][H^+]}{[\text{CH}_3\text{COOH}]} \]

Where:

• \([\text{CH}_3\text{COO}^-] \approx [H^+]\)
• \([\text{CH}_3\text{COOH}] \approx C - [H^+]\)

Substituting \([H^+] = 1.26 \times 10^{-3} \text{ M}\), you derive

• \[ 1.8 \times 10^{-5} = \frac{(1.26 \times 10^{-3})^2}{C - 1.26 \times 10^{-3}} \]

Through simple algebra, this allows us to solve for the initial concentration \(C\) of acetic acid in vinegar.

Equilibrium Expression

Chemical equilibrium occurs when the rate of the forward reaction (forming products) equals the rate of the backward reaction (reforming reactants). For acetic acid in vinegar, the chemical equilibrium involves its partial dissociation into acetate and hydrogen ions.
The equilibrium expression \(K_a\) for acids like acetic acid is given by the ratio of the product concentrations to the reactant concentrations:

• \[ K_a = \frac{[\text{CH}_3\text{COO}^-][H^+]}{[\text{CH}_3\text{COOH}]} \]

In the problem, \(K_a = 1.8 \times 10^{-5}\) indicates that acetic acid is a weak acid, meaning it doesn't completely dissociate in solution.
This weak dissociation is why we can assume \([\text{CH}_3\text{COO}^-] \approx [H^+]\), simplifying our calculations.
The concept of equilibrium helps in predicting how the reaction shifts when conditions change, maintaining the delicate balance in this reversible reaction.