Question

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.

Short answer

Based on the given information, we've found the concentrations of the different substances involved in bromoacetic acid ionization to be \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]=0.0132\: M\), \(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]=0.0868\: M\), and the \(K_{a}\) for bromoacetic acid is \(1.99\times10^{-3}\).

Step-by-step solution

Find the concentrations of \(\mathrm{H}^{+}\) and \(\mathrm{BrCH}_{2}\mathrm{COO}^{-}\)

Given that the solution is 13.2% ionized, this means that 13.2% of bromoacetic acid molecules are dissociated into \(\mathrm{H}^{+}\) and \(\mathrm{BrCH}_{2}\mathrm{COO}^{-}\). So, we can calculate the concentrations of these ions: \(\mathrm{Concentration \: of \: H}^{+}=\mathrm{Concentration \: of \: BrCH}_{2}\mathrm{COO}^{-}=\) \((0.100\: M)(0.132)=0.0132\: M\)

Find the concentration of \(\mathrm{BrCH}_{2}\mathrm{COOH}\)

Since we know the initial concentration of bromoacetic acid (0.100 M) and the concentration of dissociated \(\mathrm{H}^{+}\) and \(\mathrm{BrCH}_{2}\mathrm{COO}^{-}\) ions (0.0132 M), we can calculate the remaining concentration of \(\mathrm{BrCH}_{2}\mathrm{COOH}\): \(\mathrm{Concentration\: of\: BrCH}_{2}\mathrm{COOH}=(\mathrm{Initial\: concentration\: of\: BrCH}_{2}\mathrm{COOH})-(\mathrm{Concentration\: of\: H}^{+})\) \(\mathrm{Concentration\: of\: BrCH}_{2}\mathrm{COOH}=(0.100\: M)-(0.0132\: M)=0.0868\: M\)

Calculate \(K_{a}\)

Now, we can use the concentrations found in steps 1 and 2 to calculate \(K_{a}\) using the expression for bromoacetic acid: \(K_{a}=\frac{[\mathrm{H}^{+}][\mathrm{BrCH}_{2}\mathrm{COO}^{-}]}{[\mathrm{BrCH}_{2}\mathrm{COOH}]}\) \(K_{a}=\frac{(0.0132\: M)(0.0132\: M)}{(0.0868\: M)}=1.99\times10^{-3}\) So the \(K_{a}\) of bromoacetic acid is \(1.99\times10^{-3}\). To summarize the results: \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]=0.0132\: M\) \(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]=0.0868\: M\) \(K_{a}=1.99\times10^{-3}\)

Key concepts

Acid Dissociation Constant

The acid dissociation constant, abbreviated as \(K_a\), is a quantitative measure of an acid's strength. In simpler terms, \(K_a\) indicates how easily an acid donates a proton (\(H^+\)) when dissolved in water. For a weak acid like bromoacetic acid \(\left(\mathrm{BrCH}_2\mathrm{COOH}\right)\), not all molecules dissociate. The \(K_a\) is calculated using the established chemical equation at equilibrium:

\[\mathrm{BrCH}_2\mathrm{COOH} \rightleftharpoons {H}^+ + \mathrm{BrCH}_2\mathrm{COO}^-\]
Here, \(K_a\) is derived using the concentrations of the products \( [H^+]\) and \( [\mathrm{BrCH}_2\mathrm{COO}^-]\) divided by the concentration of the un-dissociated acid \( [\mathrm{BrCH}_2\mathrm{COOH}]\). Higher \(K_a\) values indicate stronger acids capable of donating protons more readily. The \(K_a\) can be affected by the molecular structure of the acid, such as the ability of the acidic group to stabilize the negative charge after losing its proton.

Molar Concentration

Molar concentration, often represented by \(M\), is the amount of substance, measured in moles, present in one liter of solution. It's a way to express the concentration of a solute in a solution. In the context of bromoacetic acid ionization, we consider the initial molar concentration of the acid in the solution and then determine the change in concentrations once the acid dissociates into ions. For instance, an initial concentration of 0.100 M bromoacetic acid partially ionizes, resulting in different concentrations of \(H^+\) ions and the conjugate base \(\mathrm{BrCH}_2\mathrm{COO}^-\) ions.

In calculations, molar concentration plays a crucial role as it factors into the determination of other related values, such as the acid dissociation constant \(K_a\) and percent ionization, allowing us to understand the extent of the chemical reaction.

Percent Ionization

Percent ionization describes the extent to which an acid dissociates in a solution and is a key concept in understanding the behavior of weak acids like bromoacetic acid. It's a ratio represented in percentage form, indicating how much of the original acid has turned into its ionic forms. To calculate percent ionization, we compare the concentration of the dissociated \(H^+\) ions to the initial concentration of the acid before ionization.

For example, a 13.2% ionization of a 0.100 M solution of bromoacetic acid means that 13.2% of the original acid molecules have donated their \(H^+\) ions to the solution. This percentage helps students visualize how much of the acid actually ionizes and is integral to understanding the concept of weak acid behavior in aqueous solutions.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products over time. This balance is dynamic, meaning that the reactions continue to occur, but since they happen at the same rate, the amounts stay consistent.

In the context of bromoacetic acid ionization, chemical equilibrium involves the dissociation of bromoacetic acid molecules into \(H^+\) ions and \(\mathrm{BrCH}_2\mathrm{COO}^-\) ions, and the reassociation of these ions back into bromoacetic acid. The point at which these two processes are happening at equal rates is the point of chemical equilibrium. Understanding equilibrium is crucial as it allows us to use the concentrations of the substances at equilibrium to calculate important values such as the acid dissociation constant \(K_a\) and to predict the behavior of chemical reactions.