Problem 181 - Chemistry An Atoms First Approach

Question

Consider the species \(\mathrm{PO}_{4}^{3-}, \mathrm{HPO}_{4}^{2-},\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} .\) Each ion can act as a base in water. Determine the \(K_{\mathrm{b}}\) value for each of these species. Which species is the strongest base?

Step-by-Step Solution

Answer

The \(K_{\mathrm{b}}\) values for the given ions are: \(K_{\mathrm{b}}(\mathrm{PO}_{4}^{3-}) = 1.6 \times 10^{-7}\), \(K_{\mathrm{b}}(\mathrm{HPO}_{4}^{2-}) = 2.3 \times 10^{-8}\), and \(K_{\mathrm{b}}(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}) = 1.3 \times 10^{-12}\). Among these ions, \(\mathrm{PO}_{4}^{3-}\) is the strongest base since it has the highest \(K_{\mathrm{b}}\) value (\(1.6 \times 10^{-7}\)).

Step 1: Identify the conjugate acids of each ion

When each ion acts as a base, it accepts a hydrogen ion (H+) from water. When H+ is added to each ion, we will have the following conjugate acids: 1. \(\mathrm{PO}_{4}^{3-} + \mathrm{H}^{+} \rightarrow \mathrm{HPO}_{4}^{2-}\) 2. \(\mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) 3. \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\)

Step 2: Find the \(K_{\mathrm{a}}\) values of the conjugate acids

Now we need to find the \(K_{\mathrm{a}}\) values for the conjugate acids \(\mathrm{HPO}_{4}^{2-}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\). These values can be found in a table of acid dissociation constants or in a textbook. For this exercise, we will use the following \(K_{\mathrm{a}}\) values: 1. \(K_{\mathrm{a}}(\mathrm{HPO}_{4}^{2-}) = 6.2 \times 10^{-8}\) 2. \(K_{\mathrm{a}}(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}) = 4.4 \times 10^{-7}\) 3. \(K_{\mathrm{a}}(\mathrm{H}_{3} \mathrm{PO}_{4}) = 7.5 \times 10^{-3}\)

Step 3: Calculate \(K_{\mathrm{b}}\) values for each ion

Now that we have the \(K_{\mathrm{a}}\) values for the conjugate acids, we can calculate the \(K_{\mathrm{b}}\) values for our ions using the relationship: \[K_{\mathrm{a}} K_{\mathrm{b}} = K_{\mathrm{w}}\] 1. For \(\mathrm{PO}_{4}^{3-}\): \(K_{\mathrm{b}}(\mathrm{PO}_{4}^{3-}) =\cfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}(\mathrm{HPO}_{4}^{2-})} = \cfrac{1.0 \times 10^{-14}}{6.2 \times 10^{-8}} = 1.6 \times 10^{-7}\) 2. For \(\mathrm{HPO}_{4}^{2-}\): \(K_{\mathrm{b}}(\mathrm{HPO}_{4}^{2-}) =\cfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}(\mathrm{H}_{2} \mathrm{PO}_{4}^{-})} = \cfrac{1.0 \times 10^{-14}}{4.4 \times 10^{-7}} = 2.3 \times 10^{-8}\) 3. For \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\): \(K_{\mathrm{b}}(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}) =\cfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}(\mathrm{H}_{3} \mathrm{PO}_{4})} = \cfrac{1.0 \times 10^{-14}}{7.5 \times 10^{-3}} = 1.3 \times 10^{-12}\)

Step 4: Determine the strongest base

We can determine which species is the strongest base by comparing their \(K_{\mathrm{b}}\) values. A higher \(K_{\mathrm{b}}\) value indicates a stronger base. Based on our calculations: 1. \(K_{\mathrm{b}}(\mathrm{PO}_{4}^{3-}) = 1.6 \times 10^{-7}\) 2. \(K_{\mathrm{b}}(\mathrm{HPO}_{4}^{2-}) = 2.3 \times 10^{-8}\) 3. \(K_{\mathrm{b}}(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}) = 1.3 \times 10^{-12}\) The highest \(K_{\mathrm{b}}\) value is for \(\mathrm{PO}_{4}^{3-}\) (\(1.6 \times 10^{-7}\)), so among the given ions, \(\mathrm{PO}_{4}^{3-}\) is the strongest base.

Key Concepts

Conjugate Acids and Bases

In the context of acid-base chemistry, conjugate acids and bases play a vital role in understanding chemical reactions and equilibrium. A conjugate acid is formed when a base gains a proton (H extsuperscript{+}), and conversely, a conjugate base is formed when an acid loses a proton. Let's review this concept with the ions

• Phosphate (\[\mathrm{PO}_4^{3-}\]),
• hydrogen phosphate (\[\mathrm{HPO}_4^{2-}\]),
• dihydrogen phosphate (\[\mathrm{H}_2\mathrm{PO}_4^{-}\]).

When each acts as a base, each one accepts a proton, forming their respective conjugate acids:

• Phosphate combines with H extsuperscript{+} to form \[\mathrm{HPO}_4^{2-}\],
• hydrogen phosphate becomes \[\mathrm{H}_2\mathrm{PO}_4^{-}\],
• while dihydrogen phosphate transforms into phosphoric acid (\[\mathrm{H}_3\mathrm{PO}_4\]).

This exchange of protons is the backbone of acid-base reactions, and thus, understanding the relationship between acids and their conjugate bases is crucial for predicting chemical behavior in equilibrium states.

Dissociation Constants

Dissociation constants, particularly acid dissociation constants (\[K_{ ext{a}}\]), are pivotal in understanding the strength of acids.The \[K_{ ext{a}}\] value reflects the extent to which an acid can donate a proton to water.For the conjugate acids of our species:

• Hydrogen phosphate (\[\mathrm{HPO}_4^{2-}\]) has a\[K_{ ext{a}}\] of \[6.2 \times 10^{-8}\],
• Dihydrogen phosphate (\[\mathrm{H}_2\mathrm{PO}_4^{-}\]) has a\[K_{ ext{a}}\] of \[4.4 \times 10^{-7}\],
• while phosphoric acid (\[\mathrm{H}_3\mathrm{PO}_4\]) has a higher \[K_{ ext{a}}\] of \[7.5 \times 10^{-3}\].

These values allow us to compute the \[K_{ ext{b}}\] for the bases (\[K_{ ext{a}} \times K_{ ext{b}} = K_{ ext{w}} = 1.0 \times 10^{-14}\]), directly indicating the base's strength.Calculations of \[K_{ ext{b}}\] using the known \[K_{ ext{a}}\] can thus determine how strongly a base can accept a proton, reflecting its basicity.

Chemical Equilibrium

Chemical equilibrium is an essential concept to grasp in acid-base equilibria.It occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to a stable mixture of reactants and products. In terms of acids and bases, equilibrium involves the delicate balance between dissociation and recombination of protons in the solution.The phosphate equilibrium system involves these ions \[\mathrm{PO}_4^{3-},\mathrm{HPO}_4^{2-}, \mathrm{H}_2\mathrm{PO}_4^{-}\] and their interactions with water, showing how each can achieve equilibrium by either releasing or accepting protons.In equilibrium, both the acid and its conjugate base exist in significant quantities, which means an understanding of the conjugate pairs and their associated equilibrium constants is key.This dual presence in water demonstrates the reversible nature of acid-base reactions, fitting into the larger framework of chemical equilibrium dynamics.

Base Strength Comparison

Base strength comparison hinges on the \[K_{ ext{b}}\] values – a higher \[K_{ ext{b}}\] denotes a stronger base. Given the ions provided:

• \[\mathrm{PO}_4^{3-}\] has a \[K_{ ext{b}}\] of \[1.6 \times 10^{-7}\],
• \[\mathrm{HPO}_4^{2-}\] is \[2.3 \times 10^{-8}\],
• while \[\mathrm{H}_2\mathrm{PO}_4^{-}\] is significantly less basic with a \[K_{ ext{b}}\] of \[1.3 \times 10^{-12}\].

The comparison clearly identifies \[\mathrm{PO}_4^{3-}\] as the strongest base. This is because it more readily accepts protons in solution compared to the other ions. Understanding how to compare the strength of bases can facilitate comprehension of broader chemical systems, helping predict reaction directions, buffer compositions, and the capacity of solutions to resist pH changes.