Chapter 7: Problem 90
The dissociation constant of a weak acid \(\mathrm{H} A\) is \(4.9 \times 10^{8}\). After making the necessary approximations, calculate (i) percentage ionization, (ii) \(\mathrm{pH}\) and (iii) \(\mathrm{OH}^{-}\)concentration in a decimolar solution of the acid. Water has a pH of 7 .
Short Answer
Expert verified
Percentage ionization is 0.07%, pH is 4.15, and \([OH^-]\) concentration is \(1.43 \times 10^{-10}\text{ M}\).
Step by step solution
01
Understanding the Given Data
The problem provides the dissociation constant (\(K_a\)) for a weak acid \(HA\) as \(K_a = 4.9 \times 10^{-8}\) and mentions a decimolar solution, which implies the initial concentration \([HA]_0 = 0.1 \text{ M}\).
02
Setting Up the Dissociation Equation
The dissociation of \(HA\) in water can be written as: \(HA \rightleftharpoons H^+ + A^-\). From this, the expression for the equilibrium constant is \(K_a = \frac{[H^+][A^-]}{[HA]}\).
03
Making Approximations for Weak Acid
Since \(HA\) is a weak acid, we assume \([HA]_{eq} \approx [HA]_0 = 0.1 \), and \([H^+] = [A^-] = x\) where \(x\) is the amount dissociated. Therefore, \(K_a = \frac{x^2}{0.1-x} \approx \frac{x^2}{0.1}\).
04
Calculating \([H^+]\) using Approximation
Solve for \(x\) (\([H^+]\)) using the equation: \(4.9 \times 10^{-8} = \frac{x^2}{0.1}\). This simplifies to \(x^2 = 4.9 \times 10^{-9}\), so \(x = \sqrt{4.9 \times 10^{-9}} \approx 7.0 \times 10^{-5}\text{ M}\).
05
Calculating Percentage Ionization
Percentage ionization is given by \(\frac{x}{0.1} \times 100\%\). With \(x = 7.0 \times 10^{-5}\), percentage ionization is \(\frac{7.0 \times 10^{-5}}{0.1} \times 100\% \approx 0.07\%\).
06
Calculating pH
The \(\text{pH}\) is calculated using \(\text{pH} = -\log[H^+]\). Thus, \(\text{pH} = -\log(7.0 \times 10^{-5}) \approx 4.15\).
07
Calculating \([OH^-]\) Concentration
Using the water dissociation constant \(K_w = 1.0 \times 10^{-14}\), find \([OH^-]\), where \([OH^-] = \frac{K_w}{[H^+]}\). Therefore, \([OH^-] = \frac{1.0 \times 10^{-14}}{7.0 \times 10^{-5}} \approx 1.43 \times 10^{-10}\text{ M}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dissociation Constant
The dissociation constant, abbreviated as \( K_a \), is a crucial parameter for understanding the behavior of weak acids in solution. It is a measure of the extent to which an acid can donate protons to the solution. For weak acids, \( K_a \) values are typically small. In our exercise, the weak acid \( HA \) has a dissociation constant of \( 4.9 \times 10^{-8} \). This small \( K_a \) suggests that \( HA \) does not dissociate significantly in water.
This dissociation can be expressed by the equilibrium equation:
This dissociation can be expressed by the equilibrium equation:
- \( HA \rightleftharpoons H^+ + A^- \)
- \( K_a = \frac{[H^+][A^-]}{[HA]} \)
Percentage Ionization
Percentage ionization is the fraction of the original acid that has dissociated in solution, expressed as a percentage. It's an important concept because it quantifies how much of the weak acid is dissociated into ions at a given concentration. For the problem provided, the percentage ionization helps us understand how effective the weak acid is at ionizing under certain conditions.
The formula for percentage ionization is:
The formula for percentage ionization is:
- \( \text{Percentage Ionization} = \left(\frac{[H^+]}{[HA]_0}\right) \times 100\% \)
- \( \text{Percentage Ionization} = \left(\frac{7.0 \times 10^{-5}}{0.1}\right) \times 100\% \approx 0.07\% \)
pH Calculation
Calculating the pH of a solution is a fundamental skill in chemistry, particularly when working with weak acids. The pH is derived from the concentration of hydrogen ions in the solution and is a measure of its acidity or alkalinity. For the hydrogen ion concentration \([H^+]\) obtained as \(7.0 \times 10^{-5}\), the pH is calculated using the formula:
- \( \text{pH} = -\log([H^+]) \)
- \( \text{pH} = -\log(7.0 \times 10^{-5}) \approx 4.15 \)