Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The normality of \(0.3 \mathrm{M}\) phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is a. \(0.1\) b. \(0.3\) c. \(0.6\) d. \(0.9\)

Short Answer

Expert verified
The normality of 0.3 M phosphorous acid is 0.6, so the correct answer is c.

Step by step solution

01

Understand the Concept of Normality

Normality is the measure of concentration equivalent to the gram-equivalents of a solute per liter of solution. It depends on the number of reactive units per molecule of solute, known as the equivalence factor.
02

Identify the Ionizable Protons

Phosphorous acid (\( \mathrm{H}_{3}\mathrm{PO}_{3} \)) can donate protons (\( \mathrm{H}^+ \)). However, only two of its hydrogens are efficiently ionizable. Therefore, the equivalence factor of \( \mathrm{H}_{3}\mathrm{PO}_{3} \) is 2.
03

Calculate the Normality

Use the formula: \[ \text{Normality (N)} = \text{Molarity (M)} \times \text{Equivalence Factor} \] Substitute the known values: \( 0.3 \mathrm{M} \times 2 = 0.6 \mathrm{N} \).
04

Choose the Correct Option

Compare the calculated normality with the given options. The normality of the solution is \( 0.6 \), which corresponds to option c.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Phosphorus Acid
When we talk about phosphorus acid, we're referring to its chemical composition, which is represented as \( \mathrm{H}_{3}\mathrm{PO}_{3} \). It is an example of an oxyacid, where phosphorus is bonded to oxygen and hydrogen atoms. However, it is important to distinguish that, compared to its sibling, phosphoric acid \( \mathrm{H}_{3}\mathrm{PO}_{4} \), phosphorus acid behaves differently in reactions.

The structure of phosphorus acid consists of three hydrogens, two of which are directly bonded to oxygen. In solution, only those two hydrogen atoms are prone to ionization, making them reactive as they release \( \mathrm{H}^+ \) ions. The third hydrogen is attached directly to phosphorus and does not ionize easily when dissolved in water. This behavior results in a special reactivity profile for phosphorus acid, which influences its chemical interactions and its value as a precursor in chemical syntheses.

It's worth noting that the unique properties of phosphorus acid, such as partial ionization and reactivity, make it a significant component in the agricultural and industrial sectors.
Equivalence Factor
The equivalence factor is a pathway to understanding normality in chemistry. It's a crucial concept that defines how many reactive particles or units a compound can donate, accept, or replace in a chemical reaction. When determining the equivalence factor of a substance, we look at the number of ionizable entities within one molecule of the substance.

For phosphorus acid \( \mathrm{H}_{3}\mathrm{PO}_{3} \), although it contains three hydrogen atoms, not all contribute to ionization. As discussed, only two hydrogens are ionizable, which means they can dissociate and contribute to reactivity in solutions. Therefore, the equivalence factor for phosphorus acid is 2.

Understanding the equivalence factor is essential when calculating the normality of a solution, providing insights into the number of moles of reactive entities involved.
Molarity
Molarity is a basic yet crucial concept in chemistry, representing the concentration of a solute in a solution. It is defined as the number of moles of solute present in one liter of solution. The unit of molarity is expressed in moles per liter (\( \mathrm{mol/L} \)).

When considering a solution of phosphorus acid, such as a \( 0.3 \mathrm{M} \) concentration, this means there are \( 0.3 \) moles of phosphorus acid dissolved in every liter of solution.

Molarity is not only important for mixing solutions but also serves as a basis for calculating other concentration measures, like normality. Normality is reliant on molarity and the equivalence factor, especially when acids and bases are involved. Understanding molarity allows chemists to tailor solution concentrations accurately for desired chemical reactions or analyses.
Ionization
Ionization is the process whereby molecules split to form ions, typically when dissolved in water. For acidic solutions, it's the dissociation of hydrogen ions \( \mathrm{H}^+ \) that primarily characterizes ionization.

In the case of phosphorus acid, ionization is only partial. Although phosphorus acid contains three hydrogen atoms, only two of those protons will effectively dissociate in solution to form ions \( \left(\mathrm{H}^+\right) \). This partial ionization significantly affects the acid's chemical behavior and its role in chemical reactions.

Understanding the ionization properties of phosphorus acid is crucial for determining the equivalence factor, and thus, the compound's normality when making solutions. Awareness about ionization not only helps predict and balance chemical equations but also optimizes the concentration needed for practical applications in various fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{H}_{3} \mathrm{BO}_{j}\), Boric acid is a. Aprotic acid b. dibasic acid c. tribasic acid d. monobasic acid

Like many metals, aluminium also reacts with a halogen to give a metal halide \(2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Br}_{2}(1) \rightarrow \mathrm{Al}_{2} \mathrm{Br}_{6}(\mathrm{~s})\) What quantity of \(\mathrm{Br}_{2}\) (in grams) is required for complete reaction with \(2.56 \mathrm{~g}\) of \(\mathrm{Al} ?\) What mass of white, solid \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) is expected respectively? a. \(\mathrm{Br}_{2}=11.2 \mathrm{gm}, \mathrm{Al}_{2} \mathrm{Br}_{6}=25.3 \mathrm{gm}\) b. \(\mathrm{Br}_{2}=22.7 \mathrm{gm}, \mathrm{Al}_{2} \mathrm{Br}_{6}=15.3 \mathrm{gm}\) c. \(\mathrm{Br}_{2}=22.7 \mathrm{gm}, \mathrm{Al}_{2} \mathrm{Br}_{6}=25.3 \mathrm{gm}\) d. \(\mathrm{Br}_{2}=11.2 \mathrm{gm}, \mathrm{Al}_{2} \mathrm{Br}_{6}=50.6 \mathrm{gm}\)

Which has maximum number of atoms? a. \(24 \mathrm{~g}\) of \(\mathrm{C}(12)\) b. \(56 \mathrm{~g}\) of \(\mathrm{Fe}(56)\) c. \(27 \mathrm{~g}\) of \(\mathrm{Al}(27)\) d. \(108 \mathrm{~g}\) of \(\mathrm{Ag}(108)\)

What volume of \(0.955 \mathrm{M} \mathrm{HCl}\), in milliliters, is needed to titrate \(2.152 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to the equivalence point? \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$ a. \(4.25 \mathrm{ml}\) b. \(21.5 \mathrm{ml}\) c. \(84.0 \mathrm{ml}\) d. \(42.5 \mathrm{ml}\)

The weight of a single atom of oxygen is a. \(5.057 \times 10^{23} \mathrm{~g}\) b. \(1.556 \times 10^{25} \mathrm{~g}\) c. \(2.656 \times 10^{-25} \mathrm{~g}\) d. \(4.538 \times 10^{-23} \mathrm{~g}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free