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Chapter 18: Problem 46

\(4.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is present in \(100 \mathrm{~mL}\) of the solution. What is the normality of the solution?

Short Answer

Expert verified
The normality of the solution is 1 N.

Step by step solution

01

Determine the Equivalent Weight of H鈧係O鈧�

The molecular weight of sulfuric acid (H鈧係O鈧�) is calculated by adding the atomic masses of each component: - Hydrogen (H): 1.01 g/mol 脳 2 = 2.02 g/mol - Sulfur (S): 32.07 g/mol 脳 1 = 32.07 g/mol - Oxygen (O): 16.00 g/mol 脳 4 = 64.00 g/mol So, the molar mass of H鈧係O鈧� is 2.02 + 32.07 + 64.00 = 98.09 g/mol. Since H鈧係O鈧� is a dibasic acid (releases 2 moles of H鈦� per mole of acid), the equivalent weight is 98.09 g/mol 梅 2 = 49.045 g/eq.
02

Calculate the Number of Equivalents

The number of equivalents is calculated using the weight of the solute and its equivalent weight:\[\text{Number of equivalents} = \frac{\text{mass of substance (g)}}{\text{equivalent weight (g/eq)}}\]Given that the mass of H鈧係O鈧� is 4.9 g and its equivalent weight is 49.045 g/eq, we have:\[\text{Number of equivalents} = \frac{4.9}{49.045} \approx 0.1 \text{ eq}\]
03

Calculate the Normality of the Solution

Normality (N) is defined as the number of equivalents of solute per liter of solution:\[N = \frac{\text{number of equivalents}}{\text{volume of solution (L)}}\]The given volume of the solution is 100 mL, which is equivalent to 0.1 L. Therefore:\[N = \frac{0.1 \text{ eq}}{0.1 \text{ L}} = 1 \text{ N}\]

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Key Concepts

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

Equivalent Weight
The equivalent weight of a substance is a crucial concept in chemistry. It refers to the mass of a compound that combines with or displaces either one mole of hydrogen atoms or its equivalent. This concept helps in understanding how different substances chemically interact.
To find the equivalent weight, you need the molar mass of the compound and the number of units it exchanges in a reaction. In the case of sulfuric acid (H鈧係O鈧�), it releases 2 hydrogen ions (H鈦�) in solution, classifying it as a dibasic acid.
This knowledge allows you to determine the equivalent weight of H鈧係O鈧� by dividing its molar mass (98.09 g/mol) by the number of moles of H鈦� ions it can donate, which is 2. Thus, the equivalent weight of H鈧係O鈧� becomes 49.045 g/equivalent. This value is essential when calculating normality, which is a measure of concentration in terms of equivalents.
Sulfuric Acid
Sulfuric acid is one of the most important chemicals in the industry. It is a strong mineral acid with the chemical formula H鈧係O鈧�. In the context of this problem, understanding its chemical structure and properties is necessary for calculations involving concentration and normality.
H鈧係O鈧� is known for its ability to release two hydrogen ions into solution, making it a powerful acid. Each molecule has two hydrogen atoms, one sulfur atom, and four oxygen atoms. These components contribute to its molar mass of 98.09 g/mol.
Due to its dibasic nature, H鈧係O鈧� plays a significant role in reactions where it acts as a reactant that donates hydrogen ions. Understanding this behavior makes it easier to compute essential values such as equivalent weight and subsequently determine the normality of solutions where sulfuric acid is dissolved.
Dibasic Acid
A dibasic acid is an acid that can donate two protons or hydrogen ions (H鈦�) per molecule when dissolved in water. This characteristic plays a crucial role in how the acid behaves in chemical reactions, as it can influence both the equivalent weight and the normality of a solution.
Sulfuric acid (H鈧係O鈧�) is a classic example of a dibasic acid. When it dissociates in water, it releases two moles of H鈦� ions per mole of H鈧係O鈧�, thereby doubling its impact on any acid-base reaction it participates in.
Recognizing a dibasic acid is important for calculating equivalent weight, as it requires dividing the molar mass by the number of hydrogen ions it can release. Consequently, this also affects the calculation of normality, which is based on the equivalents of solute per volume of solution. Understanding dibasic acids helps in accurately measuring and predicting the acid's strength and reactivity in various chemical contexts.

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Most popular questions from this chapter

From the following information, identify \((A)\) and \((B):\) $$ (A)+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow(B) $$ a colourless irritating gas \({ }^{\circ}(B)+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{\circ}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow\) Green soln. (a) \(\mathrm{Cl}^{-}, \mathrm{HCl}\) (b) \(\mathrm{CO}_{3}^{2-}, \mathrm{CO}_{2}\) (c) \(\mathrm{Br}^{-}, \mathrm{HBr}\) (d) \(\mathrm{SO}_{3}^{2-}, \mathrm{SO}_{2}\)

Which of the following is/are primary standard? (a) \(\mathrm{KMnO}_{4}\) (b) KOH (c) \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

Identify the incorrect statement/s. (a) The equivalent mass of \(\mathrm{KMnO}_{4}\) in alkaline medium is one third of its molecular mass. (b) The mass of oxalic acid dihydrate required to prepare \(500 \mathrm{~mL}\) of \(0.02 \mathrm{~N}\) solution is \(6.3 \mathrm{~g} .\) (c) The volume of \(0.1 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution required to oxidise \(35 \mathrm{~mL}\) of \(0.5-\mathrm{M} \mathrm{FeSO}_{4}\) solution-in acidic medium is \(29.17 \mathrm{~mL}\) (d) Phenolphthalein can be used as an indicator in iodometry.

Chlorine is prepared by reacting \(\mathrm{HCl}\) with \(\mathrm{MnO}_{2}\). The reaction is represented by the following equation : \("\) $$ \mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{Cl}_{2}+\mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ Assuming the reaction goes to completion, what mass of conc. \(\mathrm{HCl}\) solution ( \(36 \%\) HCl by mass) is needed to produce \(7.0 \mathrm{~g}\) of \(\mathrm{Cl}_{2} ?\) (a) \(20 \mathrm{~g}\) (b) \(30 \mathrm{~g}\) (c) \(40 \mathrm{~g}\) (d) \(50 \mathrm{~g}\)

(A) \(\mathrm{CdS}\) and \(\mathrm{As}_{2} \mathrm{~S}_{3}\) both are yellow coloured precipitates. (R) CdS is insoluble in yellow ammonium sulphide while \(\mathrm{As}_{2} \mathrm{~S}_{3}\) is soluble.
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