Question

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7. Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).

Short answer

The \(K_b\) for the butyrate ion is approximately \(5.0 \times 10^{-10}\).

Step-by-step solution

Note the Relationship Between pKa and pKb

Recognize that for the butyric acid (HBut), the provided value is \(\mathrm{p}K_{a} = 4.7\). The fundamental relationship between \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) is given by the formula:\[\mathrm{p}K_a + \mathrm{p}K_b = 14\]

Calculate pKb

Using the equation from Step 1, we can now calculate \(\mathrm{p}K_b\) for the conjugate base, which is the butyrate ion (But \(^-\)). Substitute the given \(\mathrm{p}K_a\) value into the equation:\[4.7 + \mathrm{p}K_b = 14\]Now, solve for \(\mathrm{p}K_b\):\[\mathrm{p}K_b = 14 - 4.7 = 9.3\]

Convert pKb to Kb

To find \(K_b\) from \(\mathrm{p}K_b\), use the formula:\[K_b = 10^{-\mathrm{p}K_b}\]Substitute \(\mathrm{p}K_b = 9.3\) into the equation:\[K_b = 10^{-9.3} \approx 5.0 \times 10^{-10}\]

Key concepts

pKa and pKb relationship

The relationship between \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) is an important concept in understanding acid-base chemistry. It involves the equation \(\mathrm{p}K_a + \mathrm{p}K_b = 14\), which connects the acid and its conjugate base. This equation is derived from the ion product of water \((K_w)\), where \(K_w = 10^{-14}\).
This means if you know the \(\mathrm{p}K_a\) of an acid, you can easily calculate the \(\mathrm{p}K_b\) of its conjugate base, and vice versa.

• A lower \(\mathrm{p}K_a\) indicates a stronger acid.
• A lower \(\mathrm{p}K_b\) indicates a stronger base.

Understanding this balance helps predict the ionization behaviour of substances in water.

conjugate base

When an acid donates a proton (H\(^+\)), it forms a conjugate base. For butyric acid (HBut), the conjugate base is butyrate ion (But\(^{-}\)).
This concept is vital in acid-base equilibria because:

• The strength of a conjugate base is inversely related to the strength of its acid.
• If the acid is strong, its conjugate base will be weak, and vice versa.

By knowing or calculating the strength of an acid, you can deduce the properties of the corresponding conjugate base.

ionization constant

The ionization constant (K) refers to how much an acid or base ionizes in solution. For acids, this is the acid dissociation constant \(K_a\), and for bases, it is \(K_b\).
Knowing the \(K_b\) of a base tells us about its strength.

• If \(K_b\) is large, the base is strong.
• If \(K_b\) is small, the base is weak.

In this exercise, we calculated \(K_b\) for butyrate ion using its \(\mathrm{p}K_b\), where \(K_b = 10^{-\mathrm{p}K_b}\). This reflects how well the base can accept a proton.

chemistry calculations

Chemistry calculations often involve using logarithmic relations, like in this exercise. Calculating \(K_b\) involved:

• Using \(\mathrm{p}K_a\) to find \(\mathrm{p}K_b\).
• Converting \(\mathrm{p}K_b\) to the ionization constant \(K_b\).

These calculations require understanding the logarithmic scale, where small changes in \(\mathrm{p}K_b\) result in large changes in \(K_b\).
This skill is crucial for predicting how different substances behave in chemical reactions and helps in designing solutions in both laboratory and industrial settings.