Problem 54
Question
A \(0.100 \mathrm{M}\) 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.
Step-by-Step Solution
Verified Answer
The concentrations of the ions and the undissociated bromoacetic acid are:
\(\left[\mathrm{H}^{+}\right] = 0.0132\,\mathrm{M}\)
\(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right] = 0.0132\,\mathrm{M}\)
\(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right] = 0.0868\,\mathrm{M}\)
\(\text{and the }K_{a}\text{ for bromoacetic acid is:}\)
\(K_{a} = 2.0 \times 10^{-3}\)
1Step 1: Determine the initial concentration of bromoacetic acid
The initial concentration is given as \(0.100\,\mathrm{M}\).
2Step 2: Calculate the concentration of dissociated bromoacetic acid
We are given that the solution is 13.2% ionized. We can calculate the concentration of dissociated bromoacetic acid by multiplying the initial concentration by the percent ionization:
\(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right] = 0.100\,\mathrm{M} \times 0.132 = 0.0132\,\mathrm{M}\)
3Step 3: Calculate the concentration of undissociated bromoacetic acid
Since the solution is 13.2% ionized, we can say that it still contains 86.8% of the initial bromoacetic acid. We can calculate the concentration of the undissociated bromoacetic acid by multiplying the initial concentration by the remaining percent:
\(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right] = 0.100\,\mathrm{M} \times 0.868 = 0.0868\,\mathrm{M}\)
4Step 4: Calculate the concentration of hydrogen ions
Since the concentration of hydrogen ions is equal to the concentration of bromoacetate ions, the concentration of hydrogen ions is:
\(\left[\mathrm{H}^{+}\right] = 0.0132\,\mathrm{M}\)
5Step 5: Calculate the \(K_{a}\) of bromoacetic acid
Using the ionization equation of a weak acid, we can calculate the \(K_{a}\) as follows:
\(K_{a} = \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]}{\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]}\)
Substitute the values obtained in Steps 2, 3 and 4 into the above equation:
\(K_{a} = \frac{0.0132\,\mathrm{M}\times 0.0132\,\mathrm{M}}{0.0868\,\mathrm{M}}= 2.0 \times 10^{-3}\)
Thus, the concentrations of the ions and the undissociated bromoacetic acid are:
\(\left[\mathrm{H}^{+}\right] = 0.0132\,\mathrm{M}\)
\(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right] = 0.0132\,\mathrm{M}\)
\(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right] = 0.0868\,\mathrm{M}\)
and the \(K_{a}\) for bromoacetic acid is:
\(K_{a} = 2.0 \times 10^{-3}\)
Key Concepts
Weak Acid IonizationPercent IonizationEquilibrium ConcentrationsChemical Equilibrium Calculations
Weak Acid Ionization
Weak acids only partially dissociate in water. This means they don’t completely break apart into ions. Understanding weak acid ionization is key to predicting how acid solutions behave. When bromoacetic acid is added to water, it only ionizes a little.
This is because weak acids like bromoacetic acid establish an equilibrium between ionized and non-ionized forms. In this case, the equilibrium consists of hydrogen ions \((\mathrm{H}^{+})\), bromoacetate ions \((\mathrm{BrCH}_{2}\mathrm{COO}^{-})\), and undissociated bromoacetic acid \((\mathrm{BrCH}_{2}\mathrm{COOH})\).
Since the degree of ionization impacts the acidity of the solution, knowing how weak acids ionize is essential to doing calculations involving acids.
This is because weak acids like bromoacetic acid establish an equilibrium between ionized and non-ionized forms. In this case, the equilibrium consists of hydrogen ions \((\mathrm{H}^{+})\), bromoacetate ions \((\mathrm{BrCH}_{2}\mathrm{COO}^{-})\), and undissociated bromoacetic acid \((\mathrm{BrCH}_{2}\mathrm{COOH})\).
Since the degree of ionization impacts the acidity of the solution, knowing how weak acids ionize is essential to doing calculations involving acids.
Percent Ionization
The percent ionization of an acid solution tells us what fraction of the original acid molecules ionized. In our example, the bromoacetic acid solution is 13.2% ionized.
This means if we started with a specific amount of bromoacetic acid, only 13.2% of it would dissociate into ions. This percentage is calculated by the formula:
This means if we started with a specific amount of bromoacetic acid, only 13.2% of it would dissociate into ions. This percentage is calculated by the formula:
- Percent Ionization = \( \frac{\text{Ionized Concentration}}{\text{Initial Concentration}} \times 100\% \)
Equilibrium Concentrations
Once a weak acid like bromoacetic acid is added to water, it reaches equilibrium between the ionized and non-ionized forms. The concentrations at equilibrium are critical for various calculations.
In our scenario, we calculated the concentrations at equilibrium for each species:
In our scenario, we calculated the concentrations at equilibrium for each species:
- The concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) is 0.0132 M.
- The concentration of bromoacetate ions \(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]\) is also 0.0132 M.
- The concentration of undissociated bromoacetic acid \(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]\) is 0.0868 M.
Chemical Equilibrium Calculations
Chemists use equilibrium concentrations to calculate other important values, like the acid dissociation constant, \( K_{a} \). This constant quantifies the strength of an acid in a solution. The equilibrium expression for calculating \( K_{a} \) of bromoacetic acid is:
This tiny value reaffirms that bromoacetic acid is a weak acid, as weak acids have lower \( K_{a} \) values. Mastering these calculations helps in assessing acid strength and predicting reaction outcomes.
- \( K_{a} = \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]}{\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]} \)
This tiny value reaffirms that bromoacetic acid is a weak acid, as weak acids have lower \( K_{a} \) values. Mastering these calculations helps in assessing acid strength and predicting reaction outcomes.
Other exercises in this chapter
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