Chapter 6: Problem 48
Write the \(K_{\mathrm{b}}\) reaction of \(\mathrm{CN}^{-}\). Given that the \(K_{\mathrm{a}}\) value for \(\mathrm{HCN}\) is \(6.2 \times 10^{-10}\), calculate \(K_{\mathrm{b}}\) for \(\mathrm{CN}^{-}\).
Short Answer
Expert verified
The \(K_{\mathrm{b}}\) for \(\mathrm{CN}^{-}\) is \(1.61 \times 10^{-5}\).
Step by step solution
01
Write the Base Ionization Reaction
The base ionization reaction for the cyanide ion, \( \mathrm{CN}^{-} \), in water can be represented as follows: \[ \mathrm{CN}^{-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^{-} \] This reaction shows the cyanide ion, \( \mathrm{CN}^{-} \), accepting a proton from water, thereby acting as a base.
02
Use the Relationship Between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\)
For a conjugate acid-base pair in water, the product of the acid dissociation constant \(K_{\mathrm{a}}\) and the base dissociation constant \(K_{\mathrm{b}}\) is equal to the ion product constant of water \(K_{\mathrm{w}}\). Therefore, \[ K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \] Where \( K_{\mathrm{w}} = 1.0 \times 10^{-14} \) at room temperature.
03
Solve for \(K_{\mathrm{b}}\)
Given that \(K_{\mathrm{a}} = 6.2 \times 10^{-10} \) for \( \mathrm{HCN} \), substitute the values into the equation: \[ (6.2 \times 10^{-10}) \times K_{\mathrm{b}} = 1.0 \times 10^{-14} \] Solve for \(K_{\mathrm{b}}\) by dividing both sides by \(6.2 \times 10^{-10}\): \[ K_{\mathrm{b}} = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.61 \times 10^{-5} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conjugate Acid-Base Pair
In the realm of chemistry, a conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton. When a base gains a hydrogen ion, it becomes its conjugate acid, whereas an acid becomes its conjugate base when it loses a hydrogen ion.
- Take the example of \[\mathrm{HCN}\] and \[\mathrm{CN}^{-}\], they form a conjugate acid-base pair. \[\mathrm{HCN}\] is the acid as it donates a proton to form \[\mathrm{CN}^{-}\], the base.
- During base ionization, the base, \[\mathrm{CN}^{-}\], accepts a proton to become \[\mathrm{HCN}\], its conjugate acid.
Ion Product Constant of Water
The ion product constant of water, often denoted as \( K_{\mathrm{w}} \), is a crucial constant in aqueous chemistry. It represents the equilibrium constant for the self-ionization of water:\[ \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^+ + \mathrm{OH}^- \]
The value of \( K_{\mathrm{w}} \) at room temperature (\( 25^\circ \mathrm{C} \)) is \( 1.0 \times 10^{-14} \). This value implies that the concentration of \( \mathrm{H}^+ \) ions multiplied by the concentration of \( \mathrm{OH}^- \) ions is constant in pure water.
The value of \( K_{\mathrm{w}} \) at room temperature (\( 25^\circ \mathrm{C} \)) is \( 1.0 \times 10^{-14} \). This value implies that the concentration of \( \mathrm{H}^+ \) ions multiplied by the concentration of \( \mathrm{OH}^- \) ions is constant in pure water.
- This constant is used in the relationship \( K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \) to calculate the unknown dissociation constants.
- For a strong acid or base, the \( K_{\mathrm{w}} \) shifts, ensuring that the ion balance in water remains stable.
Acid Dissociation Constant
The acid dissociation constant, \( K_{\mathrm{a}} \), measures the strength of an acid in a solution. It provides us with an idea of how completely an acid can donate protons to the base it reacts with. \( K_{\mathrm{a}} \) values are determined from the equilibrium of the reaction:\[ \mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^- \]
- Large \( K_{\mathrm{a}} \) values suggest strong acids, as these dissociate completely in water.
- Small \( K_{\mathrm{a}} \) values indicate weak acids, like \( \mathrm{HCN} \) which has a \( K_{\mathrm{a}} \) of \( 6.2 \times 10^{-10} \).
Base Dissociation Constant
The base dissociation constant, \( K_{\mathrm{b}} \), is similar to the acid dissociation constant but used for bases. It measures the degree to which a base will add a proton to form its conjugate acid in water. The equilibrium reaction typically looks like:\[ \mathrm{B} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BH}^+ + \mathrm{OH}^- \]
By using the relationship \( K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \), one can find the \( K_{\mathrm{b}} \) for a base if the \( K_{\mathrm{a}} \) of its conjugate acid is known.
By using the relationship \( K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \), one can find the \( K_{\mathrm{b}} \) for a base if the \( K_{\mathrm{a}} \) of its conjugate acid is known.
- The \( K_{\mathrm{b}} \) for \( \mathrm{CN}^{-} \) as derived is \( 1.61 \times 10^{-5} \), indicating its strength in gaining protons.
- Larger \( K_{\mathrm{b}} \) values typically suggest stronger bases.