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Chapter 22: Problem 62

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
H3PO3 is a diprotic acid because only two of its hydrogens are bonded to the central phosphorus atom in a P-OH group which can ionize and donate protons. Nitric acid is a strong acid because it completely ionizes in solution, while phosphoric acid is weak due to its partial ionization. Phosphate rock is ineffective as a fertilizer because the calcium phosphate minerals it contains are insoluble in water. Phosphorus exists as P4 molecules at room temperature, not as diatomic molecules like nitrogen, due to its single-bonded tetrahedral structure. Solutions of Na3PO4 are basic because the phosphate ion, being the conjugate base of the weak phosphoric acid, can accept H+ ions from water, increasing the concentration of hydroxide ions.

Step by step solution

01

Observation (a): H3PO3 is a diprotic acid.

A diprotic acid is an acid that can donate two protons (H+ ions) in an aqueous solution. Although H3PO3 (phosphorous acid) has three hydrogen atoms, it is diprotic because only two of its hydrogens are bonded to the central phosphorus (P) atom in a P-OH group. These P-OH groups can ionize and donate protons (H+) to the solution. The third hydrogen atom is directly bonded to the P atom in a P-H bond, which is not acidic.
02

Observation (b): Nitric acid is a strong acid, whereas phosphoric acid is weak.

Nitric acid (HNO3) is a strong acid because it completely ionizes or dissociates into H+ ions and nitrate ions (NO3-) in solution. This is due to the high polarity of the O-H bond and the high stability of the resultant NO3- ion. On the other hand, phosphoric acid (H3PO4) is weak because it does not dissociate completely in solution. It undergoes partial ionization and only donates one proton to the solution as it forms a dihydrogen phosphate ion (H2PO4-). The O-H bonds in phosphoric acid are not as polar as those in nitric acid, and the formed ions are not as stable, which makes phosphoric acid weaker.
03

Observation (c): Phosphate rock is ineffective as a phosphate fertilizer.

Phosphate rock is mostly composed of calcium phosphate minerals, including fluorapatite and hydroxyapatite. These minerals are insoluble in water, and plants cannot absorb nutrients from insoluble compounds. To be effective as a fertilizer, phosphate must be in a soluble form, such as ammonium phosphate or single superphosphate.
04

Observation (d): Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does.

Phosphorus and nitrogen are both non-metallic elements and exist as molecules at room temperature. However, they have different molecular structures. Nitrogen exists as a diatomic molecule (N2), which consists of two nitrogen atoms with very strong triple covalent bonds (N鈮). This triple bond is very difficult to break, making N2 a stable molecule at room temperature. In contrast, phosphorus exists as a much larger molecule, P4, at room temperature. The P4 molecule is composed of four phosphorus atoms that form a tetrahedron held together by single bonds. This structure is less stable than the N2 molecule, and hence, phosphorus does not exist as diatomic molecules at room temperature.
05

Observation (e): Solutions of Na3PO4 are quite basic.

Sodium phosphate (Na3PO4) is a salt that is formed from the reaction between a weak acid (phosphoric acid, H3PO4) and a strong base (sodium hydroxide, NaOH). When Na3PO4 is dissolved in water, it dissociates into sodium ions (Na+) and phosphate ions (PO4^3-). Since the phosphate ion is the conjugate base of the weak phosphoric acid, it can accept H+ ions from water. Therefore, it acts as a base and raises the concentration of hydroxide ions (OH-) in the solution, making the solution basic.

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Most popular questions from this chapter

Carbon forms an unusual unstable oxide of formula \(\mathrm{C}_{3} \mathrm{O}_{2}\), called carbon suboxide. Carbon suboxide is made by using \(\mathrm{P}_{2} \mathrm{O}_{5}\) to dehydrate the dicarboxylic acid called malonic acid, which has the formula \(\mathrm{HOOC}-\mathrm{CH}_{2}-\mathrm{COOH}\). (a) Write a balanced reaction for the production of carbon suboxide from malonic acid. (b) How many grams of carbon suboxide could be made from \(20.00 \mathrm{~g}\) of malonic acid? (c) Suggest a Lewis structure for \(\mathrm{C}_{3} \mathrm{O}_{2}\) - (Hint: The Lewis structure of malonic acid suggests which atoms are connected to which.) (d) By using the information in Table 8.5, predict the \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{C}_{3} \mathrm{O}_{2}\). (e) Sketch the Lewis structure of a product that could result by the addition of \(2 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to \(1 \mathrm{~mol}\) of \(\mathrm{C}_{3} \mathrm{O}_{2}\) -

Write a balanced equation for the preparation of \(\mathrm{H}_{2}\) using (a) \(\mathrm{Mg}\) and an acid, (b) carbon and steam, (c) methane and steam.

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the high-temperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\).

Write a balanced equation for each of the following reactions: (a) Hydrogen cyanide is formed commercially by passing a mixture of methane, ammonia, and air over a catalyst at \(800^{\circ} \mathrm{C}\). Water is a by-product of the reaction. (b) Baking soda reacts with acids to produce carbon dioxide gas. (c) When barium carbonate reacts in air with sulfur dioxide, barium sulfate and carbon dioxide form.

Name the following compounds and assign oxidation states to the halogens in them: (a) \(\mathrm{KClO}_{3}\), (b) \(\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}\), (c) \(\mathrm{AlCl}_{3}\), (d) \(\mathrm{HBrO}_{3}\) (e) \(\mathrm{H}_{5} \mathrm{IO}_{6}\) (f) \(\mathrm{XeF}_{4}\).
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