The molar conductance of a \(0.01 \mathrm{M}\) solution of acetic acid was found to be \(16.30 \Omega^{-1} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). The ionic conductances of hydrogen and acetate ions at infinite dilution are \(349.8\) and \(40.9 \Omega^{-1}\) \(\mathrm{cm}^{-1} \mathrm{~mol}^{-1}\), respectively, at the same temperature. What percentage of acetic acid is dissociated at this concentration? (a) \(0.04172 \%\) (b) \(4.172 \%\) (c) \(41.72 \%\) (d) \(0.4172 \%\)

Short Answer

Expert verified
4.172%

Step by step solution

01

- Calculate Limiting Molar Conductivity

Combine the ionic conductances of the hydrogen and acetate ions to find the limiting molar conductivity (\r\(\Lambda_m^\circ\)) of acetic acid. \r\[\Lambda_m^\circ = \Lambda_{+}^\circ + \Lambda_{-}^\circ\] \rFor hydrogen ion (\r\(\Lambda_{+}^\circ\)), it is \r\(349.8\ \r\Omega^{-1} \r\mathrm{cm}^{-1} \r\mathrm{mol}^{-1}\), and for acetate ion (\r\(\Lambda_{-}^\circ\)), it is \r\(40.9 \ \r\Omega^{-1} \r\mathrm{cm}^{-1} \r\mathrm{mol}^{-1}\).
02

- Calculate Degree of Dissociation

Using the molar conductance of the acetic acid solution (\r\(\Lambda_m\)) and the limiting molar conductivity (\r\(\Lambda_m^\circ\)), calculate the degree of dissociation (\r\(\alpha\)) of acetic acid. \r\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\] \rThe molar conductance \r\(\Lambda_m\) is given as \r\(16.30\ \r\Omega^{-1} \r\mathrm{cm}^{-1} \r\mathrm{mol}^{-1}\).
03

- Calculate Percentage Dissociation

To find the percentage of dissociation, multiply the degree of dissociation (\r\(\alpha\)) by 100. \r\[\text{Percentage of dissociation} = \alpha \times 100\%\]

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.

Degree of Dissociation
The concept of the degree of dissociation is central in understanding the behavior of electrolytes in a solution. It denotes the fraction of solute molecules that split into constituent ions. For weak electrolytes like acetic acid, not all molecules dissociate, leading to a value less than 1 for the degree of dissociation, symbolized by the Greek letter \( \alpha \). Mathematically, it's calculated as the ratio of the molar conductance of the electrolyte at a given concentration (\( \Lambda_m \)) to the limiting molar conductivity (\( \Lambda_m^\circ \)). The formula used is:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \]
In simpler terms, by understanding the degree of dissociation, we get insights into the extent to which a substance dissociates into ions, which is a measure of its strength as an electrolyte. This value is crucial for various applications in chemistry, including determining reaction mechanisms, calculating equilibrium constants, and in the field of electrochemistry.
Limiting Molar Conductivity
Limiting molar conductivity, represented as \( \Lambda_m^\circ \), is an essential parameter in physical chemistry that reveals the conductive capacity of an ion when the concentration approaches zero and inter-ionic interactions are negligible. This theoretical value is significant because it allows chemists to compare the ionic conductivities of different electrolytes under standard conditions.
It's determined by extrapolating the plot of molar conductivity versus the square root of concentration to zero concentration. However, for ions, \( \Lambda_m^\circ \) is the sum of the conductivities of the individual ions at infinite dilution. The equation used to find the limiting molar conductivity of acetic acid is:
\[ \Lambda_m^\circ = \Lambda_{+}^\circ + \Lambda_{-}^\circ \]
The value of \( \Lambda_m^\circ \) provides a baseline for comparison and allows for the calculation of the degree of dissociation of an electrolyte when combined with the measured molar conductance.
Ionic Conductance
Ionic conductance refers to the ability of an ion to carry an electric charge through a solution. Each ion in solution has its unique conductance value, often measured at infinite dilution to ensure no interaction between ions and a stable, consistent measurement.
In the context of our exercise, we have the ionic conductances of hydrogen (\( \Lambda_{+}^\circ \)) and acetate (\( \Lambda_{-}^\circ \)) ions, given as \(349.8 \Omega^{-1} \mathrm{cm}^{-1} \mathrm{mol}^{-1}\) and \(40.9 \Omega^{-1} \mathrm{cm}^{-1} \mathrm{mol}^{-1}\), respectively. These values highlight the conductivity if each of these ions were present alone at infinite dilution. However, in an acetic acid solution, both ions contribute to overall conductivity in accordance with their mole fraction and inherent conductance.

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

A volume of \(100 \mathrm{~m}\) of a buffer of \(1 \mathrm{M}\) \(-\mathrm{NH}_{3}\) and \(1 \mathrm{M}-\mathrm{NH}_{4}^{+}\) is placed in two half-cells connected by a salt bridge. \(A\) current of \(1.5 \mathrm{~A}\) is passed through the cell for \(20 \mathrm{~min}\). If electrolysis of water takes place only and the electrode reactions are: Right: \(2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-}\) and Left: \(2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{H}^{+}+\mathrm{O}_{2}+4 \mathrm{e}\), then, the \(\mathrm{pH}\) of the (a) right electrode will increase (b) left electrode will increase (c) both electrode will increase (d) both electrode will decrease

Which one of the following statements is incorrect regarding an electrochemical cell? (a) The electrode on which oxidation takes place is called anode. (b) Anode is the negative pole. (c) The direction of the current is same as that of the direction of flow of electrons. (d) The flow of current is partly due to flow of electrons and partly due to flow of ions.

If mercury is used as a cathode during the electrolysis of an aqueous \(\mathrm{NaCl}\) solution, the ions discharged at cathode are (a) \(\mathrm{H}^{+}\) (b) \(\mathrm{Na}^{+}\) (c) \(\mathrm{OH}^{-}\) (d) \(\mathrm{Cl}^{-}\)

The same current was passed successively through solution of zinc-ammonium sulphate and nickel-ammonium sulphate rendered alkaline with ammonia. The weights of zinc and nickel deposited in a certain time were found to be \(22.89 \mathrm{~g}\) and \(20.55 \mathrm{~g}\), respectively. Given that the chemical equivalent weight of zinc is \(32.7\), what is the chemical equivalent weight of nickel? (a) \(58.71\) (b) \(29.36\) (c) \(14.39\) (d) \(36.42\)

By how much would the oxidizing power of \(\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}\) couple change if the \(\mathrm{H}^{\circ}\) ions concentration is decreased 100 times at \(25^{\circ} \mathrm{C}\) ? (a) increases by \(189 \mathrm{mV}\) (b) decreases by \(189 \mathrm{mV}\) (c) will increase by \(19 \mathrm{mV}\) (d) will decrease by \(19 \mathrm{mV}\)

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