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Account for the following observations: (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) Solutions of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) are quite basic.

Short Answer

Expert verified
H鈧働O鈧 is diprotic due to its structure; HNO鈧 is stronger than H鈧働O鈧 due to complete dissociation; phosphate rock is insoluble; P forms polyatomic P鈧, while N forms diatomic N鈧; Na鈧働O鈧 solutions are basic due to OH鈦 formation.

Step by step solution

01

Understanding why H鈧働O鈧 is a diprotic acid

H鈧働O鈧 is phosphorous acid. Its structure is HPO(OH)鈧, where two of the hydrogen atoms are bonded to oxygen atoms and can be ionized. These ionizable hydrogen atoms contribute to the substance's acidity, making H鈧働O鈧 a diprotic acid.
02

Comparing acid strength of HNO鈧 and H鈧働O鈧

Nitric acid (HNO鈧) is a strong acid because it completely dissociates in water due to the presence of a highly electronegative nitrogen atom and a delocalized negative charge over three oxygen atoms, making the release of the proton (H鈦) very feasible. Phosphoric acid (H鈧働O鈧), on the other hand, is weak because it does not ionize completely in solution; the lack of delocalization in its conjugate base makes it less stable in its ionized form.
03

Explaining the ineffectiveness of phosphate rock as fertilizer

Phosphate rock primarily contains insoluble compounds such as calcium phosphate, which do not release phosphate ions readily in the soil. Therefore, plants cannot easily absorb these phosphates to support their growth, making raw phosphate rock relatively ineffective as a fertilizer.
04

Understanding molecular forms of phosphorus and nitrogen

Phosphorus atoms prefer to form tetrahedral P鈧 molecules at room temperature due to their larger size and the preference for bonding in a polyatomic manner. In contrast, nitrogen atoms are small and capable of forming strong triple bonds, thus existing as diatomic N鈧 molecules.
05

Explaining basic nature of Na鈧働O鈧 solutions

Sodium phosphate (Na鈧働O鈧) in solution dissociates into sodium ions (Na鈦) and phosphate ions (PO鈧劼斥伝). Phosphate ions are capable of accepting protons from water, forming HPO鈧劼测伝 and OH鈦, resulting in a basic solution (due to the production of hydroxide ions, OH鈦).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diprotic Acid
Diprotic acids are acids that can donate two protons (hydrogen ions) per molecule during the process of dissociation in solution. Let's take phosphorous acid, or \( \mathrm{H}_{3} \mathrm{PO}_{3} \), as an example. Its chemical structure is represented as \( \mathrm{HPO(OH)}_2 \), which clearly indicates the presence of two ionizable hydrogen atoms attached to oxygen.

These hydrogen atoms are the ones that can dissociate in water, qualifying \( \mathrm{H}_{3} \mathrm{PO}_{3} \) as a diprotic acid. However, it's worth noting that not all hydrogens in an acid must be ionizable. The third hydrogen in \( \mathrm{H}_{3} \mathrm{PO}_{3} \) is linked directly to phosphorus and does not easily participate in ionization.

Thus, understanding the molecular arrangement and the ability of hydrogen atoms to ionize helps explain why certain acids are categorized as diprotic.
Acid Strength
The strength of an acid is determined by how completely it dissociates in water to release protons. Nitric acid \( \mathrm{HNO}_{3} \) is classified as a strong acid because it dissociates completely in water. This complete dissociation is promoted by the strong electronegative nature of nitrogen, which stabilizes the negative charge left on the nitrates as the hydrogen ions separate.

On the other hand, phosphoric acid \( \mathrm{H}_{3} \mathrm{PO}_{4} \) is considered a weak acid due to its partial dissociation in water. The conjugate base of phosphoric acid does not have delocalized charge stability as seen in nitric acid. This lacking delocalization translates to a less stable conjugate base once the acid loses a hydrogen ion, making the release less likely to happen completely.

These differences illustrate how molecular structure directly influences acid strength.
Fertilizers
Fertilizers are substances that add essential nutrients to the soil to aid plant growth. Phosphate rock is often considered for use in fertilizers, but it is not very effective in its raw form. The primary reason is that phosphate rock consists of mostly insoluble compounds like calcium phosphate.

In the soil, these insoluble phosphates do not release phosphate ions easily, which are critical for plants to absorb and utilize. Plants require soluble phosphate for their growth processes, like root development and energy transfer within cells.

To make phosphate more available, phosphate rock is often chemically treated to produce soluble forms such as superphosphate of lime or ammonium phosphate, which are more readily taken up by plants.
Molecular Structure
Molecular structure defines how atoms bond within a molecule and the resulting shape. For phosphorus and nitrogen, their distinct molecular structures lead to different forms under standard conditions. At room temperature, phosphorus exists as \( \mathrm{P}_{4} \) molecules due to its tendency to form tetrahedral structures. Each phosphorus atom wants to form three single bonds, leading to stable \( \mathrm{P}_{4} \) tetrahedra.

In contrast, nitrogen atoms are smaller and prefer to form strong triple bonds. This gives rise to diatomic nitrogen molecules \( \mathrm{N}_{2} \), where a single nitrogen triple bond provides the stability needed for existence at room temperature.

The size of atoms and their preferred bond configurations play a critical role in determining the physical state and appearance of a substance at a given temperature.
Basic Solutions
Basic solutions are characterized by an excess of hydroxide ions \( \mathrm{OH}^- \). When sodium phosphate \( \mathrm{Na}_{3} \mathrm{PO}_{4} \) dissolves in water, it breaks into sodium ions \( \mathrm{Na}^+ \) and phosphate ions \( \mathrm{PO}_{4}^{3-} \).

The phosphate ions are particularly important here because they can react with water to capture protons and generate hydroxide ions. This reaction is written as:

\[ \mathrm{PO}_{4}^{3-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{OH}^- \]As hydroxide ions are produced, the pH of the solution rises, indicating a basic solution.

Thus, the dissociation of \( \mathrm{Na}_{3} \mathrm{PO}_{4} \) and the subsequent reactions of phosphate ions help establish its basic property in solutions.

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Most popular questions from this chapter

Complete and balance the following equations: (a) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (f) \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow\)

Consider the elements \(\mathrm{Ba}, \mathrm{Na}, \mathrm{O}, \mathrm{B}, \mathrm{P},\) and \(\mathrm{Kr}\). From this list, select the element that (a) is most electronegative, (b) has the greatest metalliccharacter, \((\mathbf{c})\) most readily forms a positive ion, \((\mathbf{d})\) exhibits a maximum oxidation sate of +5 , (e) exists as monoatomic gas at room temperature, (f) has multiple allotropes.

Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.

Write the formulas for the following compounds, and indicate the oxidation state of the group 14 element or of boron in each: (a) silane, (b) germanium dichloride, (c) tin(II) sulfide, \((\mathbf{d})\) sodium borohydride, \((\mathbf{e})\) lead(IV) nitrate, \((\mathbf{f})\) tin dioxide.

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: \((\mathbf{a}) \mathrm{N}_{2} \mathrm{O}_{3},(\mathbf{b}) \mathrm{NOCl},(\mathbf{c}) \mathrm{NO}_{2} \mathrm{Cl},(\mathbf{d}) \mathrm{N}_{2} \mathrm{O}_{4}\).

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