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

A colloidal solution is subjected to an electrical field. The particles move towards anode. The coagulation of same sol is studied using \(\mathrm{NaCl}\), \(\mathrm{BaCl}_{2}\) and \(\mathrm{AICl}_{3}\) solutions. Their coagulating power should be : (1) \(\mathrm{NaCl}>\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}\) (2) \(\mathrm{BaCl}_{2}>\mathrm{AICl}_{3}>\mathrm{NaCl}\) (3) \(\mathrm{AlCl}_{3}>\mathrm{BaCl}_{2}>\mathrm{NaCl}\) (4) \(\mathrm{BaCl}_{2}>\mathrm{NaCl}>\mathrm{AlCl}_{3}\)

Short Answer

Expert verified
(3) \(\text{AlCl}_3 > \text{BaCl}_2 > \text{NaCl}\)

Step by step solution

01

- Identify the nature of particles

The particles in the colloidal solution are moving towards the anode in an electrical field, which means they are negatively charged.
02

- Select the appropriate rule

According to Hardy-Schulze rule, the coagulating power of an ion increases with the valence (charge) of the ion. Higher the valence of the counter-ion, stronger is its coagulating power.
03

- Determine valences of ions

Identify the valences of the positive ions (cations) in \(\text{NaCl}\), \(\text{BaCl}_2\), and \(\text{AlCl}_3\): \(\text{Na}^+\) has a valence of +1, \(\text{Ba}^{2+}\) has a valence of +2, and \(\text{Al}^{3+}\) has a valence of +3.
04

- Rank the coagulating powers

Based on the Hardy-Schulze rule and the valences, rank the coagulating powers: \(\text{AlCl}_3 \) with the highest valence \(+\text{3}\), followed by \(\text{BaCl}_2\) with valence \(+\text{2}\), and \(\text{NaCl}\) with valence \(+\text{1}\). Therefore, \( \text{AlCl}_3 > \text{BaCl}_2 > \text{NaCl} \).

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.

Hardy-Schulze rule
The Hardy-Schulze rule is fundamental when understanding colloidal solutions and their coagulation behavior. This rule states that the greater the valence (charge) of the ions causing coagulation, the more effective they will be in precipitating the colloid.
For example, ions with higher charges will neutralize the charges on the colloidal particles more effectively, leading to coagulation. This is because they have stronger electrostatic attraction forces.
To put this into context, consider a colloidal solution with negatively charged particles, as identified in our exercise. When adding different solutions such as \(\text{NaCl}\), \(\text{BaCl}_2\), and \(\text{AlCl}_3\), the cations (\(\text{Na}^+\), \(Ba^{2+}\), and \(Al^{3+}\)) will neutralize the colloid particles to varying degrees. Following the Hardy-Schulze rule, \(Al^{3+}\) with a valence of +3 will have the highest coagulating power, followed by \(Ba^{2+}\) and then \(Na^+\).
Coagulation power
Coagulation power refers to the ability of an electrolyte to destabilize and aggregate particles in a colloidal solution. The coagulating power is directly influenced by the charge of the ions in the electrolyte.
According to our exercise, we are looking at \(\text{NaCl}\), \(\text{BaCl}_2\), and \(\text{AlCl}_3\). Each contains cations of different valences: \(Na^+ (valence +1)\), \(Ba^{2+} (valence +2)\), and \(Al^{3+} (valence +3)\).
The higher the valence, the more powerful the ion in causing coagulation. This is because such ions can neutralize the charge on the colloidal particles more effectively. Thus, \(Al^{3+}\) has the highest coagulating power, followed by \(Ba^{2+}\), and then \(Na^+\). In essence, the higher the charge of the ion, the stronger its coagulating power.
Electrical field in colloids
When a colloidal solution is subjected to an electrical field, the charged particles will move towards the electrode of opposite charge. In our exercise, the particles are moving towards the anode. This indicates that the colloidal particles are negatively charged.
Movement of particles in an electrical field helps in studying their charge properties and behavior. It’s a useful method for identifying the type of charge on colloidal particles.
As the charged particles migrate, they can interact with oppositely charged ions, which plays a role in the coagulation process. Understanding this behavior is crucial for applications that require controlled coagulation, such as in water purification.
Valence of ions
The valence of an ion, which is its charge, is a key concept in understanding its interaction with colloids. In the context of coagulation, ions with higher valences have a stronger capability to neutralize the charges on colloidal particles.
For example, in the given exercise, \(Na^+\) has a valence of +1, \(Ba^{2+}\) has a valence of +2, and \(Al^{3+}\) has a valence of +3. The effectiveness of these ions in causing coagulation increases with the valence.
Higher valence means stronger electrostatic force to attract and neutralize the charges on the colloidal particles, leading to faster and more effective coagulation. This principle is in direct accordance with the Hardy-Schulze rule.

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

Which equilibrium can be described as an acid base reaction using the Lewis acid - base definition but not using the Bronsted- Lowry definition? (1) \(2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons 2 \mathrm{NH}_{4}^{+}+\mathrm{SO}_{4}^{2-}\) (2) \(\mathrm{NH}_{3}+\mathrm{CH}_{3} \mathrm{COOH} ?\mathrm{NH}_{4}{ }^{*}+\mathrm{CH}_{3} \mathrm{COO}^{-}\) (3) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3} \mathrm{COOH} \quad \mathrm{H}_{3} \mathrm{O}++\mathrm{CH}_{3} \mathrm{COO}^{-}\) (4) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}+4 \mathrm{NH}_{3}=\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) \(+4 \mathrm{H}_{2} \mathrm{O}\)

Consider the following statements : \(\mathrm{S}_{1}\) : For an odd function \(\mathrm{f}(\mathrm{x})\), graph of \(y=f(x)\) always passes through origin. \(\mathbf{S}_{2}\) : If \(f\) and \(g\) are two bijective function then \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is also bijective. \(\mathbf{s}_{3}:\) All points of intersection of \(y=f(x)\) and \(y=f^{-1}(x)\) lies on \(y=x\) only. State, in order, whether \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}, \mathrm{~S}_{4}\) are true or false (1) T F T (2) T T F (3) \(\mathrm{FTT}\) (4) F FF

The electrode potentials for \(\mathrm{Cu}_{\text {(aq) }}^{2 *}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{\text {(aq) }}^{*} \quad\) and \(\mathrm{Cu}_{\text {(aq) }}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}\) are \(+0.15 \mathrm{~V}\) and \(+0.50 \mathrm{~V}\) respectively. The value of \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}\) will be : (1) \(0.500 \mathrm{~V}\) (2) \(0.325 \mathrm{~V}\) (3) \(0.650 \mathrm{~V}\) (4) \(0.150 \mathrm{~V}\)

Acetaldehyde and Propyne can be distinguish by: (i) Tollen's reagent (ii) \(\mathrm{I}_{2} / \mathrm{NaOH}\) (iii) Lucas reagent (iv) neutral \(\mathrm{FeCl}_{3}\) (1) (i), (ii) \& (iii) (2) (ii) \& (iii) (3) (i) \& (ii) (4) (iii) \& (iv)

STATEMENT - 1 : If the roots of equation \(a x^{4}+b x^{3}+c x^{2}+d x+e=0\) are in H.P then the roots of the equation \(e x^{4}+d x^{3}+c x^{2}+b x+a=0\) are in A.P. STATEMENT- 2 : The equation having reciprocals of the roots of the equation \(f(x)=0\), as roots is \(f\left(\frac{1}{x}\right)=0\) (1) Statement \(-1\) is True, Statement \(-2\) is True ; Statement \(-2\) is a correct explanation for Statement \(-1\) (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement- 1 (3) Statement \(-1\) is True, Statement \(-2\) is False (4) Statement \(-1\) is False, Statement \(-2\) is True

See all solutions

Recommended explanations on English 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