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Chapter 9: Problem 54

\(50 \mathrm{ml}\) of \(10 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}, 25 \mathrm{ml}\) of \(12 \mathrm{~N} \mathrm{HCl}\) and 40 \(\mathrm{ml}\) of \(5 \mathrm{~N} \mathrm{HNO}_{3}\) are mixed and the volume of the mixture is made \(1000 \mathrm{ml}\) by adding water. The normality of resulting solution will be (a) \(1 \mathrm{~N}\) (b) \(4 \mathrm{~N}\) (c) \(3 \mathrm{~N}\) (d) \(2 \mathrm{~N}\)

Short Answer

Expert verified
The normality of the resulting solution is 1 N (option a).

Step by step solution

01

Understanding Normality

Normality (N) is a measure of concentration equivalent to equivalent per liter of solution. It can be calculated using the formula: \[ \text{Normality} = \frac{\text{Equivalents of solute}}{\text{Liters of solution}} \]
02

Calculate Equivalents for Each Acid

First, we need to find the total equivalents of each acid.**For \(H_2SO_4\):**- Volume = 50 ml = 0.050 L- Normality = 10 N- Equivalent = Normality 脳 Volume = 10 脳 0.050 = 0.5 equivalents**For \(HCl\):**- Volume = 25 ml = 0.025 L- Normality = 12 N- Equivalent = Normality 脳 Volume = 12 脳 0.025 = 0.3 equivalents**For \(HNO_3\):**- Volume = 40 ml = 0.040 L- Normality = 5 N- Equivalent = Normality 脳 Volume = 5 脳 0.040 = 0.2 equivalents
03

Sum Total Equivalents

Add the equivalents of all acids to find the total equivalents in the mixture.- Total equivalents = 0.5 (from \(H_2SO_4\)) + 0.3 (from \(HCl\)) + 0.2 (from \(HNO_3\))- Total equivalents = 1.0 equivalents
04

Calculate the Final Normality

The mixture is diluted to a total volume of 1000 ml or 1.0 L. Use the normality formula:\[ \text{Normality of solution} = \frac{\text{Total equivalents}}{\text{Total volume in L}} = \frac{1.0}{1.0} = 1.0 \text{ N} \]
05

Determine the Correct Answer

Comparing the calculated normality with the given options, the correct normality of the resulting solution is \(1 \text{ N}\). Hence, the answer is option (a) \(1 \text{ N}\).

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Key Concepts

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

Concentration of Solutions
The concentration of a solution tells us how much solute is dissolved in a given amount of solvent. It is important in chemistry because it affects various properties like reactivity and pH. One common way to express concentration is through normality (N), which represents the number of equivalents of solute per liter of solution.
  • Molecular Weight: This is typically used in grams per mole and helps calculate equivalences.
  • Equivalent Weight: The molecular weight divided by the number of equivalents. For acids, this refers to the number of hydrogen ions the acid can furnish.
  • Calculating Normality: Normality can be affected by dilution, mixing, and chemical reactions.
This leads to the need for precise calculations when solutions with different strengths or volumes are combined, especially in reactions and analysis.
Equivalents of Solute
When we talk about equivalents in the context of chemistry, we鈥檙e discussing a measure that allows calculation of amounts in reactions, considering the reactivity of acidic or basic species. The concept of equivalents hinges on the ability of a molecule to undergo a reaction with a set number of ions like hydrogen or hydroxide ions.
  • Equivalent Concept: One equivalent of an acid is the amount that reacts with one mole of hydroxide ions.
  • Finding Equivalents: For acids, it鈥檚 often calculated by multiplying volume (in liters) by the normality of the solution.
  • Use in Reactions: Knowing equivalents helps balance chemical equations and predict the outcome of chemical reactions accurately.
This is particularly beneficial when mixing solutions of varying normalities, as it allows for the calculation of total reactivity.
Dilution Effect
Dilution involves reducing the concentration of a solution by increasing its volume. Adding solvent, like water, decreases concentration while maintaining the total amount of solute present in the solution. This principle is key to safely handling strong acids or other concentrated chemicals.

The dilution equation, \( M_1V_1 = M_2V_2 \), is typically used in molarity but is also applicable when analyzing normality:
  • Initial and Final Volumes: Changes in these volumes require adjustments in concentration calculation.
  • Use in Laboratory: Allows work with safer, manageable concentrations.
  • Practical Utility: Used in adjusting the concentrations of stock solutions to needed reaction conditions.
In this exercise, diluting to a final volume of 1000 ml adjusts the combined total equivalents to form a new normality, providing a reliable way to mix and estimate concentration after chemical processing.
Acid Solution Mixing
When combining different acid solutions, normality helps us understand the final strength of the mixture. This mixing is crucial not only in lab processes but also in industrial applications where precise acid concentration is critical for processes like pH adjustment and corrosion control.

Understanding how different acids mix and react is critical because:
  • Final Normality: By summing the equivalents provided by different acids, we get an idea of the overall acidity.
  • Handling Different Acids: Each acid contributes differently based on its chemical properties and concentration.
  • Application in Reactions: Ensures that the desired acidity level is achieved for subsequent chemical reactions or even industrial processes.
Thus, the focus should be on how mixing results in a solution that is uniquely balanced, maintaining desired properties of acidity or reactivity.

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Most popular questions from this chapter

(A): Camphor is used as a solvent in the determination of molecular weight on non-volatile solute. (R): Camphor has high molal elevation constant.

Out of \(1 \mathrm{~N} \mathrm{HNO}_{3}\) and \(1 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) aqueous solutions, which has higher boiling point? a. \(\mathrm{HNO}_{3}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) c. both have same boiling point d. cannot be predicted

A \(0.001\) molal solution of a complex \(\left[\mathrm{MX}_{8}\right]\) in water has the freezing point of \(-0.0054^{\circ} \mathrm{C}\). Assuming \(100 \%\) ionization of the complex salt and \(\mathrm{K}_{f}\) for \(\mathrm{H}_{2} \mathrm{O}=1.86 \mathrm{Km}^{-1}\), write the correct representation for the complex. a. \(\left[\mathrm{MX}_{6}\right] \mathrm{X}_{2}\) b. \(\left[\mathrm{MX}_{5}\right] \mathrm{X}_{3}\) c. \(\left[\mathrm{MX}_{8}\right]\) d. \(\left[\mathrm{MX}_{7}\right] \mathrm{X}\)

A solution containing \(0.1 \mathrm{~g}\) of a non-volatile organic substance \(\mathrm{P}\) (molecular weight \(=100\) ) in \(100 \mathrm{~g}\) of benzene raises the boiling point of benzene by \(0.2^{\circ} \mathrm{C}\), while a solution containing \(0.1 \mathrm{~g}\) of another non-volatile substance \(\mathrm{Q}\) in the same amount of benzene raises the boiling point of benzene by \(0.4^{\circ} \mathrm{C}\). What is the ratio of molecular weights of \(\mathrm{P}\) and \(\mathrm{Q}\) ? a. \(1: 4\) b. \(4: 1\) c. \(1: 2\) d. \(2: 1\)

The vapour pressure of water at \(20^{\circ} \mathrm{C}\) is \(17.5 \mathrm{~mm}\) Hg. If \(18 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is added to \(178.2 \mathrm{~g}\) of water at \(20^{\circ} \mathrm{C}\), the vapour pressure of the resulting solution will be (a) \(17.675 \mathrm{~mm} \mathrm{Hg}\) (b) \(15.750 \mathrm{~mm} \mathrm{Hg}\) (c) \(16.500 \mathrm{~mm} \mathrm{Hg}\) (d) \(17.325 \mathrm{~mm} \mathrm{Hg}\).
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