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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Advanced 2018 · Paper 2 · Chemistry · Ionic Equilibrium

The dilution processes of different aqueous solutions with water...

The dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of the solutions on [H+][ {H}^{+} ] are given in LIST-II.
(Note: The degree of dissociation \left(\alpha\right)ofaweakacidandaweakbaseisof a weak acid and a weak base is< < 1;thedegreeofhydrolysisofsaltisthe degree of hydrolysis of salt is< < 1; [ {H}^{+} ]representstheconcentrationofrepresents the concentration of{H}^{+}ions)<tableborder="1"><tbody><tr><td></td><td>LIST–I</td><tdstyle="text−align:justify;"></td><tdstyle="text−align:justify;">LIST−II</td></tr><tr><td>A)</td><td>ions)<table border="1"> <tbody> <tr> <td> </td> <td>LIST –I</td> <td style="text-align: justify;"> </td> <td style="text-align: justify;">LIST -II</td> </tr> <tr> <td>A)</td> <td>10mLof0 . 1MNaOH + 20mLof0 .1 Maceticaciddilutedtoacetic acid diluted to60mL.</td><td>P)</td><td>Thevalueof.</td> <td>P)</td> <td>The value of[ {H}^{+} ]doesnotchangeondilution.</td></tr><tr><td>B)</td><td>does not change on dilution.</td> </tr> <tr> <td>B)</td> <td>20mLof0 . 1MNaOH + 20mLof0 . 1Maceticaciddilutedtoacetic acid diluted to80mL.</td><td>Q)</td><td>Thevalueof.</td> <td>Q)</td> <td>The value of{\text{[H}}^{+} ]changestohalfofitsinitialvalueondilution.</td></tr><tr><td>C)</td><td>changes to half of its initial value on dilution.</td> </tr> <tr> <td>C)</td> <td>20mLof0 .1 MHCl + 20mLof0 .1 Mammoniasolutiondilutedtoammonia solution diluted to80mL.</td><td>R)</td><td>Thevalueof.</td> <td>R)</td> <td>The value of[ {H}^{+} ]changestotwotimesitsinitialvalueondilution.</td></tr><tr><td>D)</td><td>changes to two times its initial value on dilution.</td> </tr> <tr> <td>D)</td> <td>10mLsaturatedsolutionofsaturated solution ofNi{\left(OH\right)}_{2}inequilibriumwithexcessofsolidin equilibrium with excess of solidNi{\left(OH\right)}_{2}isdilutedtois diluted to20mL(solid(solidNi{\left(OH\right)}_{2}isstillpresentafterdilution).</td><td>S)</td><td>Thevalueofis still present after dilution).</td> <td>S)</td> <td>The value of[ {H}^{+} ]changestochanges to\sqrt{}timesitsinitialvalueondilution.</td></tr><tr><td></td><td></td><td>T)</td><td>Thevalueoftimes its initial value on dilution.</td> </tr> <tr> <td> </td> <td> </td> <td>T)</td> <td>The value of[ {H}^{+} ]changestochanges to\sqrt{2}$ times its initial value on dilution.
Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option isThe dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of theThe dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of theThe dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of the

Options

  1. Aa-q;b-t;c-p;d-r;
  2. Ba-p;b-t;c-s;d-p;
  3. Ca-s;b-t;c-p;d-r;
  4. Da-s;b-q;c-p;d-t;

Correct answer

B. a-p;b-t;c-s;d-p;

Step-by-step solution

(A)( A )
The dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of the
pH=pKa⇒pH = pKa ⇒\left[{H}^{+}\right] will not change on dilution.
(B)( B )
The dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of the
\left[{OH}^{-}\right]=KHC=(kwkaC)= \sqrt{{K}_{H}C} = \sqrt{\left(\dfrac{{k}_{w}}{{k}_{a}}C\right)}<br><br >{\left[{H}^{+}\right]}_{1} = \sqrt{\dfrac{{k}_{w}{k}_{a}}{C}}<br><br >\dfrac{{\left[{H}^{+}\right]}_{2}}{{\left[{H}^{+}\right]}_{1}} = \sqrt{\dfrac{{C}_{1}}{{C}_{2}}} = \sqrt{\dfrac{0 .05}{0 .025}}= \sqrt{2}<br><br >( C )<br><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeeadvanced/yydrhsjxojftvakhhijoheucyrlcosggumziuxiwzsxmulncrn.jpeg"/><br>[H+]=KHC<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/yydrhsjxojftvakhhijoheucyrlcosggumziuxiwzsxmulncrn.jpeg"/><br >\left[{H}^{+}\right] = \sqrt{{K}_{H}C}
[H+]2[H+]1=C2C1=12\dfrac{{\left[{H}^{+}\right]}_{2}}{{\left[{H}^{+}\right]}_{1}} = \sqrt{\dfrac{{C}_{2}}{{C}_{1}}} = \dfrac{1}{\sqrt{2}}
(D)( D ) Because of dilution, the solubility does not change. So, \left[{H}^{+}\right]=constant= constant.
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