Chapter 13: Problem 14
If the \(K_{\mathrm{a}}\) for \(\mathrm{HCN}\) is \(6.2 \times 10^{-10},\) what is the \(K_{\mathrm{b}}\) for \(\mathrm{CN}^{-}(\mathrm{aq}) ?\)
Short Answer
Expert verified
The \(K_b\) for \(\text{CN}^-\) is approximately \(1.61 \times 10^{-5}\).
Step by step solution
01
Understanding the Relationship Between Ka and Kb
The problem requires us to find the base dissociation constant, \(K_b\), of the cyanide ion, \(\text{CN}^-\) from the acid dissociation constant, \(K_a\), of hydrogen cyanide, \(\text{HCN}\). According to the relation for conjugate acid-base pairs, \(K_a \times K_b = K_w\), where \(K_w\) is the ion product constant of water, which is \(1.0 \times 10^{-14}\) at room temperature.
02
Solving for Kb
To find \(K_b\), rearrange the formula: \(K_b = \frac{K_w}{K_a}\). Substitute the given \(K_a\) value into this equation: \(K_b = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}}\).
03
Calculating the Result
Divide \(1.0 \times 10^{-14}\) by \(6.2 \times 10^{-10}\) to find \(K_b\). This equates to: \[K_b = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} \approx 1.61 \times 10^{-5}.\]
04
Conclusion of Calculation
The base dissociation constant, \(K_b\), for \(\text{CN}^-\) is approximately \(1.61 \times 10^{-5}\). This confirms the relationship between the weak acid \(\text{HCN}\) and its conjugate base \(\text{CN}^-\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ka and Kb relationship
In acid-base chemistry, the relationship between the acid dissociation constant, \(K_a\), and the base dissociation constant, \(K_b\), plays a critical role in understanding the properties of acids and bases. These constants quantify the strengths of acids and bases.
In particular, they allow us to explore how easily an acid donates protons or how readily a base accepts protons. For a conjugate acid-base pair, \(K_a\) and \(K_b\) are mathematically linked by the ion product of water, \(K_w\).
To elaborate:
In particular, they allow us to explore how easily an acid donates protons or how readily a base accepts protons. For a conjugate acid-base pair, \(K_a\) and \(K_b\) are mathematically linked by the ion product of water, \(K_w\).
To elaborate:
- The equation \(K_a \times K_b = K_w\) establishes this fundamental relationship.
- Since \(K_w\) is a constant value of \(1.0 \times 10^{-14}\) at room temperature, this means that knowing either \(K_a\) or \(K_b\) allows us to determine the other.
- For example, if we know the \(K_a\) of a weak acid like HCN, we can find the \(K_b\) of its conjugate base, CN\(^{-}\), which is precisely what the original exercise accomplished.
Conjugate acid-base pairs
Conjugate acid-base pairs are a central concept in the Brønsted-Lowry theory of acids and bases. Whenever an acid donates a proton, it transforms into its conjugate base. Similarly, a base that gains a proton becomes its conjugate acid.
These pairs make it easier to track proton transfer processes in chemical reactions.
Here are some important points to remember about conjugate acid-base pairs:
These pairs make it easier to track proton transfer processes in chemical reactions.
Here are some important points to remember about conjugate acid-base pairs:
- A conjugate base arises from the deprotonation of an acid. For instance, when hydrogen cyanide (HCN) loses a proton, it forms the cyanide ion (CN\(^-\)), which acts as the conjugate base.
- Conversely, when a base gains a proton, it becomes a conjugate acid. This is observable when CN\(^-\) accepts a proton to re-form HCN.
- The strength of an acid inversely correlates with the strength of its conjugate base. A weak acid will have a comparatively stronger conjugate base, which is the case with HCN and CN\(^-\).
Ion product of water
The ion product of water, symbolized as \(K_w\), represents the product of the concentrations of the hydrogen ion, \([H^+]\), and the hydroxide ion, \([OH^-]\), in pure water.
At 25 °C (room temperature), \(K_w\) has a value of \(1.0 \times 10^{-14}\). This tiny constant reflects the fact that water is predominantly molecular, yet small amounts do dissociate into ions.
Understanding \(K_w\) is crucial because:
At 25 °C (room temperature), \(K_w\) has a value of \(1.0 \times 10^{-14}\). This tiny constant reflects the fact that water is predominantly molecular, yet small amounts do dissociate into ions.
Understanding \(K_w\) is crucial because:
- It sets the stage for the calculations in acid-base chemistry by defining how acidic or basic a solution can be purely by the dissociation of water.
- Under neutral conditions, \([H^+] = [OH^-]\), and therefore each ion has a concentration of \(1.0 \times 10^{-7}\) M.
- For an acidic solution, \([H^+] > [OH^-]\), while for a basic solution, \([OH^-] > [H^+]\).
- The value of \( K_w \) changes with temperature, which means the acidity or basicity of a solution can also depend on temperature.