/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 What will be the normality of so... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What will be the normality of solution obtained by mixing 20 \(\mathrm{mL}\) of \(0.5 \mathrm{~N}\) hydrochloric acid, \(60 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) sulphuric acid and \(150 \mathrm{~mL}\) of \(0.2 \mathrm{~N}\) nitric acid?

Short Answer

Expert verified
The normality of the solution is 0.2 N.

Step by step solution

01

Calculate the milliequivalents of hydrochloric acid

To find the milliequivalents (meq) of hydrochloric acid (HCl), use the formula: \[ ext{meq} = N imes V \] where \(N\) is the normality and \(V\) is the volume in mL. For HCl: \[ ext{meq} = 0.5 imes 20 = 10 \text{ meq} \]
02

Calculate the milliequivalents of sulphuric acid

For sulphuric acid (Hâ‚‚SOâ‚„), use the same formula: \[ ext{meq} = 0.1 imes 60 = 6 \text{ meq} \]
03

Calculate the milliequivalents of nitric acid

For nitric acid (HNO₃), use the same formula: \[ ext{meq} = 0.2 imes 150 = 30 \text{ meq} \]
04

Find the total milliequivalents in the mixture

Add the milliequivalents from each acid to find the total milliequivalents in the mixture: \[ ext{Total meq} = 10 + 6 + 30 = 46 \text{ meq} \]
05

Calculate the total volume of the mixture

Add the volumes of all solutions to find the total volume: \[ V_{ ext{total}} = 20 + 60 + 150 = 230 \text{ mL} \]
06

Calculate the normality of the mixture

The normality of the mixture is given by the formula: \[ N = \frac{ ext{Total meq}}{V_{ ext{total}}} \] Substitute the known values: \[ N = \frac{46}{230} = 0.2 \text{ N} \]

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Milliequivalents
Milliequivalents (meq) are a unit of chemical measurement commonly used in calculations involving acids and bases. They are useful because they help standardize measurements based on the number of reactive units in a substance, rather than the amount of the substance itself. This is important in chemistry when dealing with reactions or solutions.

To calculate milliequivalents, you multiply the normality of the solution (\(N\), which is the equivalent concentration of solute per liter of solution) by the volume of the solution in milliliters (\(V\)).
This gives us the equation: \( \text{meq} = N \times V \).
  • Normality (\(N\)) refers to the concentration of equivalents per liter of solution.
  • Volume (\(V\)) is the amount of the solution in milliliters.
By using milliequivalents, you are essentially counting how many reactive units are present, which is useful for determining the outcome of reactions or the strength of a solution. In our exercise, we use this method to assess the contributions of different acids in a mixture.
Mixing Different Acids
When you mix different acids, you're bringing together solutions with varied characteristics. In the exercise, we're tasked with mixing hydrochloric acid, sulphuric acid, and nitric acid.
Each of these acids has its own normality and volume, and this affects the final composition of the mixture.
To understand the process:
  • Hydrochloric acid (\(HCl\)): Starts with a normality of 0.5N and a volume of 20 mL, giving 10 meq.
  • Sulphuric acid (\(H_2SO_4\)): Has a normality of 0.1N and a volume of 60 mL, contributing 6 meq.
  • Nitric acid (\(HNO_3\)): With a normality of 0.2N and a larger volume of 150 mL, adds 30 meq to the mixture.
When these acids are combined, the total milliequivalents are calculated by summing up the contributions from each acid.
This results in a total of 46 meq. The total volume of the mixture is the sum of the individual volumes: 230 mL.
Finding the Normality Step-by-Step
To determine the normality of an acid mixture, follow a systematic step-by-step approach. This ensures accuracy and avoids confusion. Here's how it's done:
  • Calculate Milliequivalents for Each Acid: Using the formula \( \text{meq} = N \times V \), find the milliequivalents for each component in the mixture. This turns individual normality and volume into a comparable unit across all components.
  • Find Total Milliequivalents: Add up the milliequivalents you've calculated for each acid. This total represents the overall reactive capacity of the mixture.
  • Calculate Total Volume: Add the volumes of the individual acids to find the complete volume of the mixture.
  • Determine the Mixture's Normality: With total milliequivalents and total volume, calculate the final normality using \( N = \frac{\text{Total meq}}{V_{\text{total}}} \). Substitute the known values to find the normality.
In our case, substituting 46 total meq and 230 mL volume into this formula gives a normality of 0.2N.
This step-by-step method not only helps find the answer accurately but also aids in understanding the roles different variables play in the calculation.

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

How much water should be added to \(100 \mathrm{~mL}\) of \(2.5 \mathrm{NH}_{2} \mathrm{SO}_{4}\) to make it normal solution?

6 g sample of \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}+\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) mixture and an inert material was treated with excess of aqueous \(\mathrm{KI}\) which reduces all \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) ion. The resulting solution was diluted to \(50 \mathrm{~mL}\). \(10 \mathrm{~mL}\) of the diluted solution titrated with \(5.5 \mathrm{~mL}\) of \(1 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) to react with all iodine. The \(\mathrm{I}_{2}\) from another \(25 \mathrm{~mL}\) sample was extracted after which the \(\mathrm{Fe}^{2+}\) was titrated with \(3.2 \mathrm{~mL}\) of \(1 \mathrm{MMnO}_{4}^{-}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Calculate percentage of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) in the mixture. [Hint : Reactions involved are : (1) \(\mathrm{Fe}_{3} \mathrm{O}_{4}+2 \mathrm{KI}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{FeSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}\) (2) \(\mathrm{Fe}_{2} \mathrm{O}_{3}+2 \mathrm{KI}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{FeSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+3 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}\) (3) \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \Gamma+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (4) \(\left.\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\right]\). \(25 \mathrm{~m} \mathrm{~L} \mathrm{H}_{2} \mathrm{O}_{2}\) were added to excess of acidified solution of \(\mathrm{KI}\). The iodine so liberated required \(20 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) sodium

If \(40 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{KOH}\) is mixed \(60 \mathrm{~mL}\) of \(0.05 \mathrm{~N} \mathrm{HCl}\), what will be the normality of the resulting solution? What will be the normality of salt formed after the reaction?

\(0.35 \mathrm{~g}\) of a metal was dissolved in \(50 \mathrm{~mL} N\) -acid. The whole solution then required \(20.85 \mathrm{~mL}\) of normal alkaline solution to neutralise the excess of the acid. Calculate the equivalent mass of the metal.

\(30 \mathrm{~mL}\) of sodium carbonate solution is mixed with \(20 \mathrm{~mL}\) of \(0.8 \mathrm{~N}\) sulphuric acid. The resultant solution needed \(20 \mathrm{~mL}\) of \(0.7 \mathrm{~N}\) hydrochloric acid solution for complete neutralisation. Determine the strength of the sodium carbonate in grams per litre. (Take sodium carbonate to be anhydrous).

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.