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 equivalent conductance of \(\mathrm{Ba}^{2+}\) and \(\mathrm{Cl}^{-}\)are respectively 127 and \(76 \mathrm{ohm}^{-1} \mathrm{~cm}^{2}\) equiv \(^{-1}\) at infinite dilution. The equivalent conductance (in ohm \(^{-1} \mathrm{~cm}^{2}\) equiv \(^{-1}\) ) of \(\mathrm{BaCl}_{2}\) at infinite dilution will be (a) \(139.5\) (b) 203 (c) 279 (d) \(101.5\)

Short Answer

Expert verified
The equivalent conductance of \(\text{BaCl}_2\) at infinite dilution is 279 ohm\(^{-1} \text{ cm}^{2}\) equiv\(^{-1}\).

Step by step solution

01

Understanding Equivalent Conductance

Equivalent conductance at infinite dilution of an electrolyte is the sum of the equivalent conductances of the cation and anion. It is measured in units of ohm\(^{-1} \text{ cm}^{2} \text{ equiv}^{-1}\). For a compound like \(\text{BaCl}_2\), which dissociates into \(\text{Ba}^{2+}\) and 2 \(\text{Cl}^{-}\), we need to account for both ions.
02

Identify Individual Ion Contributions

We are given the equivalent conductance of \(\text{Ba}^{2+}\) as 127 \(\text{ohm}^{-1} \text{ cm}^{2} \text{ equiv}^{-1}\), and \(\text{Cl}^{-}\) as 76 \(\text{ohm}^{-1} \text{ cm}^{2} \text{ equiv}^{-1}\). Each molecule of \(\text{BaCl}_2\) provides 1 \(\text{Ba}^{2+}\) and 2 \(\text{Cl}^{-}\) ions.
03

Calculate Total Conductance of \(\text{BaCl}_2\)

The equivalent conductance of \(\text{BaCl}_2\) can be calculated by adding the equivalent conductance of 1 \(\text{Ba}^{2+}\) ion and 2 \(\text{Cl}^{-}\) ions:\[\lambda_{\text{BaCl}_2} = \lambda_{\text{Ba}^{2+}} + 2\lambda_{\text{Cl}^{-}}\]Substitute the given values:\[\lambda_{\text{BaCl}_2} = 127 + 2 \times 76 = 127 + 152\]
04

Final Calculation

Continuing from the previous step:\[\lambda_{\text{BaCl}_2} = 127 + 152 = 279\]Therefore, the equivalent conductance of \(\text{BaCl}_2\) at infinite dilution is 279 \(\text{ohm}^{-1} \text{ cm}^{2} \text{ equiv}^{-1}\).

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.

Infinite Dilution
Infinite dilution is a term used in chemistry to describe the state where an electrolyte is dissolved in a solvent to such an extent that it is completely dissociated into its ions. This state is crucial for measuring true properties of electrolytes like equivalent conductance. At infinite dilution, interactions between ions of the electrolyte are negligible due to the vast amount of solvent.
Understanding infinite dilution helps chemists determine the true conductance of an electrolyte, as it is unaffected by the intermolecular forces that might exist at higher concentrations. These interactions can skew conductance measurements by facilitating or impeding ion movement. Thus, physicists and chemists rely on the concept of infinite dilution to extract accurate intrinsic properties of ions within solutions. These measurements provide vital parameters in electrolyte chemistry and help in calculating factors such as equivalent conductance.
Electrolyte Conductivity
Electrolyte conductivity refers to the ability of ions in a solution to conduct electricity. This property is essential in understanding how solutions work, especially ionic solutions used in various technological applications. It is typically measured by assessing how well a substance allows the flow of electric current.
Conductivity is influenced by several factors:
  • Ion concentration: More ions mean higher conductivity.
  • Types of ions: Different ions carry different charges, affecting conductivity levels.
  • Temperature: Higher temperatures typically increase conductivity.
When electrolytes are dissolved in water, they break into cations and anions, which carry electric charge through the solution. By measuring the electrolyte conductivity at infinite dilution, scientists can evaluate the maximum capacity of ions to conduct electricity without ionic interactions interfering. This is extremely beneficial in industries like battery production and water purification.
Ionic Contributions
In a compound, the overall conductivity is determined by the individual contributions of each ion present. For example, in the compound \(\mathrm{BaCl}_2\), dissociation results in one \(\mathrm{Ba}^{2+}\) ion and two \(\mathrm{Cl}^-\) ions. Each ion type has its own equivalent conductance, a unique measurement highlighting its efficiency in conducting electricity within the solution.
To find the total equivalent conductance of a compound at infinite dilution, as demonstrated in the solution, the equivalent conductance of all individual ions must be summed up according to their stoichiometric proportions in the compound. Therefore:
  • Calculate the contribution of each ion.
  • Consider stoichiometry: multiply each ion’s equivalent conductance by the number of times it appears in the formula.
For \(\mathrm{BaCl}_2\), the conductance formula becomes \(\lambda_{\text{BaCl}_2} = \lambda_{\mathrm{Ba}^{2+}} + 2\lambda_{\mathrm{Cl}^-}\). This calculation reflects the combined effect of all ions in the solution, providing a complete picture of the electrolyte's conductive properties at infinite dilution.

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

When a lead storage battery is discharged (a) lead is formed (b) lead sulphate is consumed (c) \(\mathrm{SO}_{2}\) is evolved (d) sulphuric acid is consumed

In acidic medium \(\mathrm{MnO}_{4}^{-}\)is an oxidizing agent \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} .\) If \(\mathrm{H}^{+}\)ion concentration is doubled, electrode potential of the half cell \(\mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+} / \mathrm{Pt}\) will (a) increase by \(28.46 \mathrm{mV}\) (b) decrease by \(28.46 \mathrm{mV}\) (c) increase by \(14.23 \mathrm{mV}\) (d) decrease by \(142.30 \mathrm{mV}\)

\(4.5 \mathrm{~g}\) of aluminium (at. mass \(27 \mathrm{amu}\) ) is deposited at cathode from \(\mathrm{Al}^{3+}\) solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from \(\mathrm{H}^{+}\)ions is solution by the same quantity of electric charge will be (a) \(44.8 \mathrm{~L}\) (b) \(22.4 \mathrm{~L}\) (c) \(11.2 \mathrm{~L}\) (d) \(5.6 \mathrm{~L}\)

Two electrochemical cells \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) and \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) are con- nected in series. What will be the net emf of the cell at \(25^{\circ} \mathrm{C} ?\) Given: \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}=-0.73 \mathrm{~V}\), \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}=+0.34 \mathrm{~V}\) \(\mathrm{Fe}^{2+} \mid \mathrm{Fe}=-0.41 \mathrm{~V}\) (a) \(+1.85\) (b) \(-1.85 \mathrm{~V}\) (c) \(+0.83 \mathrm{~V}\) (d) \(-0.83 \mathrm{~V}\)

What is the quantity of electricity (in coulombs) required to deposit all the silver from \(250 \mathrm{~mL}\) of \(1 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) solution? \((\mathrm{Ag}=108)\) (a) \(2412.5\) (b) 24125 (c) \(4825.0\) (d) 48250

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