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Ionic Equilibrium question

2022 · 25 Jul · Shift 2 · Q3
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  5. /2022 · 25 Jul · Shift 2 · Q3

Ionic Equilibrium question

2022 · 25 Jul · Shift 2 · Q3

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Ka1K_{a_1}Ka1​​, Ka2K_{a_2}Ka2​​ and Ka3K_{a_3}Ka3​​ are the respective ionization constants for the following reactions (a), (b) and (c). (a) H2C2O4⇌H++HC2O4−\mathrm{H_2C_2O_4} \rightleftharpoons \mathrm{H^+} + \mathrm{HC_2O_4^-}H2​C2​O4​⇌H++HC2​O4−​(b) HC2O4−⇌H++C2O42−\mathrm{HC_2O_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2O_4^{2-}}HC2​O4−​⇌H++C2​O42−​(c) H2C2O4⇌2H++C2O42−\mathrm{H_2C_2O_4} \rightleftharpoons 2\mathrm{H^+} + \mathrm{C_2O_4^{2-}}H2​C2​O4​⇌2H++C2​O42−​ The relationship between Ka1K_{a_1}Ka1​​, Ka2K_{a_2}Ka2​​ and Ka3K_{a_3}Ka3​​ is given as :
  1. A
    Ka3=Ka1+Ka2{K_{{a_3}}}={K_{{a_1}}}+{K_{{a_2}}}Ka3​​=Ka1​​+Ka2​​
  2. B
    Ka3=Ka1−Ka2{K_{{a_3}}}={K_{{a_1}}}-{K_{{a_2}}}Ka3​​=Ka1​​−Ka2​​
  3. C
    Ka3=Ka1/Ka2{K_{{a_3}}}={K_{{a_1}}}/{K_{{a_2}}}Ka3​​=Ka1​​/Ka2​​
  4. D
    Ka3=Ka1×Ka2{K_{{a_3}}}={K_{{a_1}}}\times{K_{{a_2}}}Ka3​​=Ka1​​×Ka2​​
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