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JEE Main 202225 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2022 · 25 July · Evening Shift · Chemistry · Ionic Equilibrium

Ka _ 1 , Ka _ 2 and Ka _ 3 are the respective ionization constants...

{Ka}_{1}, {Ka}_{2} and {Ka}_{3}aretherespectiveionizationconstantsforthefollowingreactions(a),are the respective ionization constants for the following reactions \left(a\right),\left(b\right)and(c).</p><p>(a)and \left(c\right).</p><p>(a){H}_{2}{C}_{2}{O}_{4} ⇌ {H}^{+} + {HC}_{2}{O}_{4}^{-}</p><p>(b)</p><p>(b){HC}_{2}{O}_{4}^{-} ⇌ {H}^{+} + {HC}_{2}{O}_{4}^{2 -}</p><p>(c)</p><p>(c){H}_{2}{C}_{2}{O}_{4} ⇌ 2{H}^{+} + {C}_{2}{O}_{4}^{2 -}</p><p>Therelationshipbetween</p><p>The relationship between{K}_{{a}_{1}}, {K}_{{a}_{2}} and {K}_{{a}_{3}}isgivenas</p>is given as</p>

Options

  1. AKa3=Ka1+Ka2{K}_{a3} = {K}_{a1} + {K}_{a2}
  2. BKa3=Ka1Ka2{K}_{a3} = \dfrac{{K}_{a1}}{{K}_{a2}}
  3. CKa3=Ka1−Ka2{K}_{a3} = {K}_{a1} - {K}_{a2}
  4. DKa3=Ka1×Ka2{K}_{a3} = {K}_{a1} \times {K}_{a2}

Correct answer

D. Ka3=Ka1×Ka2{K}_{a3} = {K}_{a1} \times {K}_{a2}

Step-by-step solution

H2C2O4⇌H++HC2O4−Ka1{H}_{2}{C}_{2}{O}_{4} ⇌ {H}^{+} + {HC}_{2}{O}_{4}^{-}{K}_{{a}_{1}}

H2C2O4−⇌H++C2O42−Ka2{H}_{2}{C}_{2}{O}_{4}^{-} ⇌ {H}^{+} + {C}_{2}{O}_{4}^{2 -}{K}_{{a}_{2}}

H2C2O4⇌2H++C2O42−Ka3{H}_{2}{C}_{2}{O}_{4} ⇌ 2{H}^{+} + {C}_{2}{O}_{4}^{2 -}{K}_{{a}_{3}}

On adding equation \left(1\right)andequation(2)and equation \left(2\right) we get equation \left(3\right).So,. So,{K}_{a3} = {K}_{a1} \times {K}_{a2}.

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