Question

(a) Given that \(K_{a}\) for cyanic acid is \(3.5 \times 10^{-4}\) and that for hydrofluoric acid is \(6.8 \times 10^{-4},\) which is the stronger acid? (b) Which is the stronger base, the cyanate ion or the fluoride ion? (c) Calculate \(K_{b}\) values for \(\mathrm{NCO}^{-}\) and \(\mathrm{F}^{-}\).

Short answer

(a) Hydrofluoric acid is the stronger acid. (b) The cyanate ion (NCO^-) is the stronger base. (c) Kb values for NCO- and F- are \(2.86 \times 10^{-11}\) and \(1.47 \times 10^{-11}\), respectively.

Step-by-step solution

Determine the stronger acid

Given the Ka values for cyanic acid (\(3.5 \times 10^{-4}\)) and hydrofluoric acid (\(6.8 \times 10^{-4}\)), we can determine which one is a stronger acid. The larger the Ka value, the stronger the acid. Therefore, we can compare the Ka values: Cyanic acid Ka: \(3.5 \times 10^{-4}\) Hydrofluoric acid Ka: \(6.8 \times 10^{-4}\) From these values, we can see that hydrofluoric acid has a larger Ka value and is, therefore, the stronger acid.

Determine the stronger base

The corresponding bases for cyanic acid and hydrofluoric acid are the cyanate ion (NCO^-) and the fluoride ion (F^-), respectively. To determine which one is a stronger base, we can compare their Kb values. However, we only have their Ka values at the moment. To find the Kb values, we will use the ion-product constant for water, Kw. The relationship between the Ka, Kb and Kw can be stated as: Kw = Ka × Kb Where Kw (the ion-product constant for water) is equal to \(1.0 \times 10^{-14}\) at 25°C.

Calculate Kb values for cyanate and fluoride ions

Using the relation Kw = Ka × Kb, we'll calculate the Kb values for both NCO- and F-. First, rearrange the equation to solve for Kb: Kb = Kw / Ka For the cyanate ion (NCO^-): Kb(NCO^-) = \(1.0 \times 10^{-14} / 3.5 \times 10^{-4}\) For the fluoride ion (F^-): Kb(F^-) = \(1.0 \times 10^{-14} / 6.8 \times 10^{-4}\)

Compare Kb values and determine the stronger base

Now, let's compare the Kb values and determine which ion is the stronger base. First, calculate the numerical values: Kb(NCO^-) = \(2.86 \times 10^{-11}\) Kb(F^-) = \(1.47 \times 10^{-11}\) Since the Kb value for the cyanate ion (NCO^-) is larger, it is the stronger base compared to the fluoride ion (F^-). In summary: (a) Hydrofluoric acid is the stronger acid. (b) The cyanate ion (NCO^-) is the stronger base. (c) Kb values for NCO- and F- are \(2.86 \times 10^{-11}\) and \(1.47 \times 10^{-11}\), respectively.

Key concepts

Ka (acid dissociation constant)

The acid dissociation constant, often denoted by \(K_a\), is a measure of the strength of an acid in solution. It helps predict how readily an acid can donate a proton (\(H^+\)) to water, forming hydronium ions \([H_3O^+]\).

A higher \(K_a\) value means that the acid dissociates more completely in water, making it stronger. For example, with \(K_a\) values given for cyanic acid \(3.5 \times 10^{-4}\) and hydrofluoric acid \(6.8 \times 10^{-4}\), hydrofluoric acid is confirmed to be the stronger acid as it has the larger \(K_a\) value.

• \(K_a = [H_3O^+][A^-] / [HA]\) (where \([HA]\) is the concentration of undissociated acid)
• A strong acid has a higher \(K_a\); a weak acid has a lower \(K_a\).
• The larger the \(K_a\), the more the acid tends to release \(H^+\) in solution.

Kb (base dissociation constant)

The base dissociation constant, \(K_b\), represents how strongly a base attracts protons in solution. It's a crucial part of determining base strength. With \(K_b\), we can evaluate how effectively a base dissociates in water to accept a \(H^+\), forming hydroxide ions \([OH^-]\).

It's interconnected with \(K_a\) by the relationship: \(K_w = K_a \times K_b\), where \(K_w\) at 25°C is a constant \(1.0 \times 10^{-14}\).

• For any acid-base pair, their dissociation constants are related by \(K_w\).
• To calculate \(K_b\) from \(K_a\): \(K_b = K_w / K_a\).
• Example: For cyanate ion \(NCO^-\) with \(K_a\) of \(3.5 \times 10^{-4}\), \(K_b = 2.86 \times 10^{-11}\).
• A higher \(K_b\) indicates a stronger base.

ion-product constant (Kw)

The ion-product constant for water, \(K_w\), is a fundamental concept in acid-base chemistry.

It describes the extent of water's ability to autoionize, meaning self-ionize, into \(H^+\) and \(OH^-\) ions. At 25°C, \(K_w\) is approximately \(1.0 \times 10^{-14}\) mol²/L².

• In pure water: \([H^+] = [OH^-] = 1.0 \times 10^{-7}\) M
• \(K_w = [H^+][OH^-]\), representing water's equilibrium state.
• The \(K_w\) value remains constant at a given temperature.
• Using \(K_w\) helps relate \(K_a\) and \(K_b\) for acid-base conjugate pairs. For example, if \(K_a\) decreases, \(K_b\) must increase.

Understanding \(K_w\) is crucial for accurately assessing the properties of acids, bases, and their respective conjugate pairs.

Lewis acids and bases

Lewis acids and bases expand the traditional definition, focusing on electron-pair exchange rather than protons.

A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. This definition helps understand reactions in broader contexts, including those that do not involve \(H^+\) ions directly.

• Lewis Acid: Accepts an electron pair. Example: \(BF_3\), \(AlCl_3\).
• Lewis Base: Donates an electron pair. Example: \(NH_3\), \(OH^-\).
• This framework helps explain many chemical reactions outside of the usual acid-base paradigm.
• Allows for a more generalized view of bonding and reaction mechanisms.

By understanding Lewis acids and bases, students can grasp the nuances of complex reactions, including those with transition metals and coordination compounds.