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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​​
View written solutionFree

Correct answer: D

  1. Write the equilibrium constants for each reaction

For reaction (a):

H2C2O4⇌H++HC2O4−\mathrm{H_2C_2O_4} \rightleftharpoons \mathrm{H^+} + \mathrm{HC_2O_4^-}H2​C2​O4​⇌H++HC2​O4−​

its ionization constant is

Ka1=[H+][HC2O4−][H2C2O4]K_{a_1}=\frac{[\mathrm{H^+}][\mathrm{HC_2O_4^-}]}{[\mathrm{H_2C_2O_4}]}Ka1​​=[H2​C2​O4​][H+][HC2​O4−​]​

For reaction (b):

HC2O4−⇌H++C2O42−\mathrm{HC_2O_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2O_4^{2-}}HC2​O4−​⇌H++C2​O42−​

its ionization constant is

Ka2=[H+][C2O42−][HC2O4−]K_{a_2}=\frac{[\mathrm{H^+}][\mathrm{C_2O_4^{2-}}]}{[\mathrm{HC_2O_4^-}]}Ka2​​=[HC2​O4−​][H+][C2​O42−​]​

For reaction (c):

H2C2O4⇌2H++C2O42−\mathrm{H_2C_2O_4} \rightleftharpoons 2\mathrm{H^+} + \mathrm{C_2O_4^{2-}}H2​C2​O4​⇌2H++C2​O42−​

its ionization constant is

Ka3=[H+]2[C2O42−][H2C2O4]K_{a_3}=\frac{[\mathrm{H^+}]^2[\mathrm{C_2O_4^{2-}}]}{[\mathrm{H_2C_2O_4}]}Ka3​​=[H2​C2​O4​][H+]2[C2​O42−​]​
  1. Relate reaction (c) to reactions (a) and (b)

Reaction (c) is obtained by adding reactions (a) and (b):

H2C2O4⇌H++HC2O4−\mathrm{H_2C_2O_4} \rightleftharpoons \mathrm{H^+} + \mathrm{HC_2O_4^-}H2​C2​O4​⇌H++HC2​O4−​ HC2O4−⇌H++C2O42−\mathrm{HC_2O_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2O_4^{2-}}HC2​O4−​⇌H++C2​O42−​

Adding them, HC2O4−\mathrm{HC_2O_4^-}HC2​O4−​ cancels:

H2C2O4⇌2H++C2O42−\mathrm{H_2C_2O_4} \rightleftharpoons 2\mathrm{H^+} + \mathrm{C_2O_4^{2-}}H2​C2​O4​⇌2H++C2​O42−​

which is exactly reaction (c).

  1. Use the rule for equilibrium constants

When two reactions are added, their equilibrium constants are multiplied.

Therefore,

Ka3=Ka1×Ka2K_{a_3}=K_{a_1}\times K_{a_2}Ka3​​=Ka1​​×Ka2​​
  1. Check directly by multiplying expressions
K_{a_1}K_{a_2}= \left(\frac{[\mathrm{H^+}][\mathrm{HC_2O_4^-}]}{[\mathrm{H_2C_2O_4}]}}\right) \left(\frac{[\mathrm{H^+}][\mathrm{C_2O_4^{2-}}]}{[\mathrm{HC_2O_4^-}]}}\right)

Cancelling [HC2O4−][\mathrm{HC_2O_4^-}][HC2​O4−​]:

Ka1Ka2=[H+]2[C2O42−][H2C2O4]K_{a_1}K_{a_2}=\frac{[\mathrm{H^+}]^2[\mathrm{C_2O_4^{2-}}]}{[\mathrm{H_2C_2O_4}]}Ka1​​Ka2​​=[H2​C2​O4​][H+]2[C2​O42−​]​

But this is exactly Ka3K_{a_3}Ka3​​, so

Ka3=Ka1Ka2K_{a_3}=K_{a_1}K_{a_2}Ka3​​=Ka1​​Ka2​​
  1. Evaluate options
  • A: Ka3=Ka1+Ka2K_{a_3}=K_{a_1}+K_{a_2}Ka3​​=Ka1​​+Ka2​​ ❌
  • B: Ka3=Ka1−Ka2K_{a_3}=K_{a_1}-K_{a_2}Ka3​​=Ka1​​−Ka2​​ ❌
  • C: Ka3=Ka1/Ka2K_{a_3}=K_{a_1}/K_{a_2}Ka3​​=Ka1​​/Ka2​​ ❌
  • D: Ka3=Ka1×Ka2K_{a_3}=K_{a_1}\times K_{a_2}Ka3​​=Ka1​​×Ka2​​ ✅

Hence, the correct option is D.

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