Problem 53 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The concentrations of the dissociated and undissociated chloroacetic acid and the hydrogen ion concentration are: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right] = \left[\mathrm{H}^{+}\right] = 0.011 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right] = 0.089 M\] and the acid dissociation constant, \(K_{a}\) for chloroacetic acid is \(1.36 \times 10^{-3}\).

Step 1: 1. Calculate the concentration of \(\mathrm{ClCH}_{2}\mathrm{COO}^{-}\) and \(\mathrm{H}^{+}\)

Since we know that the \(0.100 M\) solution is \(11.0\%\) ionized, we can calculate the concentration of ions: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right] = \left[\mathrm{H}^{+}\right] = 0.100 M \times 0.110 = 0.011 M\]

Step 2: 2. Calculate the concentration of \(\mathrm{ClCH}_{2}\mathrm{COOH}\)

To calculate the concentration of the non-ionized chloroacetic acid, we need to subtract the concentrations of ions from the initial concentration: \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right] = 0.100 M - 0.011 M = 0.089 M\]

Step 3: 3. Calculate the acid dissociation constant, \(K_{a}\)

The equilibrium expression for the dissociation of chloroacetic acid is given by: \[K_{a} = \frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]}\] Substitute the values obtained in steps 1 and 2 into the expression to calculate \(K_{a}\): \[K_{a} = \frac{(0.011)(0.011)}{0.089} \approx 1.36 \times 10^{-3}\] So, the concentrations of the dissociated and undissociated chloroacetic acid and the hydrogen ion concentration are: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right] = \left[\mathrm{H}^{+}\right] = 0.011 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right] = 0.089 M\] and the acid dissociation constant, \(K_{a}\) for chloroacetic acid is \(1.36 \times 10^{-3}\).

Key Concepts

Ionization

Ionization is a crucial concept in chemistry, especially when dealing with acids and their behavior in solutions. When an acid like chloroacetic acid dissolves in water, it partially dissociates into ions. In this case, the ionization occurs as the chloroacetic acid separates into chloroacetate anion (\( \mathrm{ClCH}_{2}\mathrm{COO}^{-} \)) and hydrogen cation (\( \mathrm{H}^{+} \)). The extent to which this ionization happens is important because it influences the concentration of ions and ultimately the acid's properties in solution.For the given chloroacetic acid solution, it is noted that the ionization is only 11%. This means that out of the total acid molecules present, only a small fraction has converted into ions.

• The concentration of ions formed in the solution is directly related to the percentage ionization and the initial concentration of the acid solution.
• By multiplying the initial molarity by the percentage ionized, one can determine the concentration of each ion formed. In this case, 0.100 M of acid and 11% ionization result in 0.011 M of chloroacetate and hydrogen ions.

Understanding how ionization works can help predict how acidic or conductive a solution will be.

Chloroacetic Acid

Chloroacetic acid, known chemically as \( \mathrm{ClCH}_{2}\mathrm{COOH} \), is an organochlorine compound and a type of carboxylic acid. This acid is a key compound in organic synthesis due to its high reactivity and ability to form derivatives. Due to the presence of a chlorine atom attached to the alpha carbon, chloroacetic acid exhibits higher acidity compared to acetic acid.

• The added chlorine atom increases its acidity by stabilizing the negative charge on the chloroacetate ion that forms during dissociation.
• This stabilization occurs through an effect known as "inductive effect," wherein the electron-withdrawing nature of the chlorine atom helps spread out the charge.

With a relatively significant ionization percentage, chloroacetic acid is a moderately strong acid in water. This increased acidity makes it valuable in reactions where strong acid catalysts are required.

Equilibrium Expression

The equilibrium expression is a fundamental tool in understanding how chemical reactions behave in a state of balance. For the dissociation reaction of chloroacetic acid, the equilibrium expression provides insights into how ion concentrations relate to each other and the undissociated acid at equilibrium.The expression for the equilibrium constant (\( K_a \)) of chloroacetic acid is given by:\[ K_a = \frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]} \]

• \([\mathrm{ClCH}_{2}\mathrm{COO}^{-}]\) and \([\mathrm{H}^{+}]\) represent the concentrations of the dissociated ions.
• \([\mathrm{ClCH}_{2}\mathrm{COOH}]\) is the concentration of the acid that remains undissociated.

Inserting the given concentrations into this formula allows the calculation of \( K_a \) for chloroacetic acid, which quantifies its strength. A stronger acid will have a larger \( K_a \) value, indicating a greater tendency to donate protons. Here, the \( K_a \) is calculated as \( 1.36 \times 10^{-3} \), marking chloroacetic acid as a relatively strong acid compared to many other carboxyl acids.