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Boric acid is prepared from the borax by the action of (1) hydrochloric acid (2) sodium hydroxide (3) carbon dioxide (4) sodium carbonate

Short Answer

Expert verified
Hydrochloric acid (1)

Step by step solution

01

Identify Reaction Requirements

Determine what borax and boric acid are chemically. Borax is chemically represented as \( \text{Na}_2\text{B}_4\text{O}_7 \bullet 10\text{H}_2\text{O} \) and boric acid as \( \text{H}_3\text{BO}_3 \).
02

Understand Possible Reactions

List the possible reactions with each given reagent. For hydrochloric acid: \( \text{Na}_2\text{B}_4\text{O}_7 \bullet 10\text{H}_2\text{O} + 2\text{HCl} \rightarrow 4\text{H}_3\text{BO}_3 + 2\text{NaCl} + 5\text{H}_2\text{O} \). Check if the other reagents can also produce boric acid.
03

Verify the Correct Reaction

Review the reactions to determine which one leads to the creation of boric acid. The reaction with hydrochloric acid produces boric acid directly.
04

Select the Correct Answer

Based on the reactions, confirm that the correct reagent to produce boric acid from borax is hydrochloric acid.

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Key Concepts

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

borax
Borax is an essential mineral that you need to understand while preparing boric acid. Chemically, borax is represented as \(\text{Na}_2\text{B}_4\text{O}_7 \bullet 10\text{H}_2\text{O}\). This means it is composed of sodium, boron, oxygen, and water. Borax is often used in cleaning products, glass manufacturing, and even in the production of cosmetics. When you dissolve borax in water, it breaks down into borate ions (\(\text{B}_4\text{O}_7^{2-}\)), sodium ions (\(\text{Na}^+\)), and water molecules. This setup is critical to the chemical reaction where boric acid will be produced.
hydrochloric acid
Hydrochloric acid (HCl) is a strong acid and plays a significant role in preparing boric acid from borax. When borax reacts with hydrochloric acid according to the equation: \(\text{Na}_2\text{B}_4\text{O}_7 \bullet 10\text{H}_2\text{O} + 2\text{HCl} \rightarrow 4\text{H}_3\text{BO}_3 + 2\text{NaCl} + 5\text{H}_2\text{O}\), it produces boric acid (\text{H}_3\text{BO}_3), sodium chloride (\text{NaCl}), and water (\text{H}_2\text{O}). This reaction is straightforward and efficient because hydrochloric acid provides the necessary hydrogen ions (H\(^+\)), which bond with borate to form boric acid. This is why hydrochloric acid is chosen over other substances like sodium hydroxide or carbon dioxide.
chemical reactions
Understanding chemical reactions is pivotal for this exercise. A chemical reaction involves the transformation of reactants into products. In preparing boric acid, the reaction between borax and hydrochloric acid is an excellent example. The reactants—borax and hydrochloric acid—undergo a chemical change to produce boric acid, sodium chloride, and water. The equation: \(\text{Na}_2\text{B}_4\text{O}_7 \bullet 10\text{H}_2\text{O} + 2\text{HCl} \rightarrow 4\text{H}_3\text{BO}_3 + 2\text{NaCl} + 5\text{H}_2\text{O}\) demonstrates this transformation clearly. Recognizing the formation, breakdown, and rearrangement of bonds helps in predicting product formation in chemical reactions.
engineering entrance exams
Grasping the concept of boric acid preparation from borax is important for engineering entrance exams. These exams test your understanding of fundamental chemical reactions and processes. Knowing the correct reagent, in this case, hydrochloric acid, showcases your ability to apply theoretical knowledge practically. Engineering exams often include similar problems requiring you to identify proper reagents, predict products of reactions, and balance chemical equations. These concepts are the building blocks of more complex topics in chemistry and engineering. Practicing such exercises diligently can improve your problem-solving skills and confidence during exams.

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Most popular questions from this chapter

Boron differs from the other clements of its own group in several properties. The anomalous behaviour of boron may be attributed to (1) its small size and high ionization energy. (2) its high electronegativity. (3) the presence of only two electrons in the penultimate shell and the absence of \(d\) orbitals in the valence shell. (4) All of the above.

Boron differs from the other elements of the group because it has (1) a different configuration in its penultimate shell. (2) a valence shell electronic configuration of \(s^{2} p^{1} .\) (3) \(10 d\) -electrons in third orbit. (4) high ionization potential.

In the reaction of aluminium with dilute sodium hydroxide solution, which of the following is not correct? (1) Hydrogen gas will be liberated. (2) Sodium meta aluminate is formed. (3) The product exists as \(\left|\mathrm{Al}(\mathrm{OH})_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right|\). (4) Co-ordination number of aluminium in the product is 4 .

Identify the false statement regarding diborane. (1) It has B-B covalent bond. (2) Boron atom is involved in \(s p^{3}\) hybridization. (3) It has a maximum of six planar atoms. (4) It contains two \(3 \mathrm{c}-2 \mathrm{e}\) bonds.

Which of the following statements is not correct regarding the structure of diboranc? (1) Therc are two bridging hydrogen atoms in diboranc. (2) Each boron atom forms four bonds in diborane. (3) \Lambdall the hydrogen atoms are not in the same planc in diboranc. (4) \Lambdall B II bonds in diborane are similar.

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