Problem 151 - A Complete Resource Book in Chemistry for JEE Main

Question

At what concentration of \(\mathrm{CH}_{3} \mathrm{COOH}\) will the \(\left[\mathrm{H}^{+}\right]\) obtained will be same as that obtained from \(10^{-2} \mathrm{M}\) \(\mathrm{HCOOH},\left(\mathrm{Ka}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=10^{-5}, \mathrm{Ka}(\mathrm{HCOOH})=10^{-4}\right)\) (a) \(10 \mathrm{M}\) (b) \(5 \mathrm{M}\) (c) \(10^{-1} \mathrm{M}\) (d) \(6 \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of CH₃COOH is 0.1 M (option c).

Step 1: Calculate the Hydronium Ion Concentration for HCOOH

Given that the concentration of HCOOH is \(10^{-2} \text{ M}\) and \( \text{Ka(HCOOH)} = 10^{-4} \). Since HCOOH is a weak acid, we can use the formula:\[[\text{H}^+] = \sqrt{\text{Ka} \times C}\]where \(C\) is the concentration of the acid. Therefore,\[[\text{H}^+] = \sqrt{10^{-4} \times 10^{-2}} = \sqrt{10^{-6}} = 10^{-3} \text{ M}\].

Step 2: Set up the Equation for CH3COOH

We want \([\text{H}^+]\) for \(\text{CH}_3\text{COOH}\) to be the same as that of HCOOH. Using the Ka expression for \(\text{CH}_3\text{COOH}\), we have:\[[\text{H}^+] = \sqrt{\text{Ka(CH}_3\text{COOH)} \times C}\].We know \([\text{H}^+] = 10^{-3} \text{ M}\).

Step 3: Solve for the Concentration of CH3COOH

Substitute the values into the equation:\[10^{-3} = \sqrt{10^{-5} \times C}\].Squaring both sides gives\[10^{-6} = 10^{-5} \times C\].Solving for \(C\), we get:\[C = \frac{10^{-6}}{10^{-5}} = 10^{-1} \text{ M}\].

Key Concepts

Acid Dissociation Constant (Ka)

The acid dissociation constant, denoted as \( K_a \), is a crucial concept when dealing with weak acids. It measures the strength of an acid in solution. The larger the \( K_a \), the stronger the acid; meaning it dissociates more in the solution. A weak acid only partially dissociates to release hydronium ions \( [\mathrm{H}^+] \).

For example, in the exercise we see acetic acid \( (\mathrm{CH}_3\mathrm{COOH}) \) with a \( K_a \) of \( 10^{-5} \) and formic acid \( (\mathrm{HCOOH}) \) with a \( K_a \) of \( 10^{-4} \). It's evident that formic acid is a stronger acid compared to acetic acid because it has a higher \( K_a \). Understanding these constants can help us predict the behaviour of acids in various scenarios, making it easier to anticipate their strengths and compositions in solutions.

Hydronium Ion Concentration

The concentration of hydronium ions \( [\mathrm{H}^+] \) in a solution is key to understanding acidity. It indicates how many protons (or hydrogen ions) are present, which defines the pH level of the solution.

In weak acid solutions, the hydronium ion concentration is calculated using the formula \([\mathrm{H}^+] = \sqrt{K_a \times C}\), where \( C \) is the initial concentration of the acid. This formula exemplifies the partial dissociation typical of weak acids.

For instance, in the problem, given formic acid \( (\mathrm{HCOOH}) \) concentration is \( 10^{-2} \text{ M} \), the calculation \([\mathrm{H}^+] = \sqrt{10^{-4} \times 10^{-2}} = 10^{-3} \text{ M}\) derives the hydronium ion concentration, showing that even at a robust initial concentration, the solution remains only moderately acidic due to the limited dissociation.

Weak Acid Concentration Calculations

Calculating the concentration of weak acids entails working backward from the desired hydronium ion concentration in many cases. Once the required \( [\mathrm{H}^+] \) is known, you can determine the necessary initial concentration of a weak acid using its \( K_a \).

For example, to match the formic acid solution's \( [\mathrm{H}^+] \) of \( 10^{-3} \text{ M} \) with acetic acid, we set up the equation \([\mathrm{H}^+] = \sqrt{K_a(\mathrm{CH}_3\mathrm{COOH}) \times C}\). With \( K_a \) given for acetic acid, squaring both sides and solving for \( C \) gives \( C = \frac{10^{-6}}{10^{-5}} = 10^{-1} \text{ M} \). This calculation shows how varying the initial concentration compensates for differing acid strengths, allowing solution acidity to be tailored as needed.