Medium MCQ +4 / -1 PYQ · JEE Mains 2022

${K_{{a_1}}}$, ${K_{{a_2}}}$ and ${K_{{a_3}}}$ are the respective ionization constants for the following reactions (a), (b) and (c).

(a) $${H_2}{C_2}{O_4} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + H{C_2}O_4^ - $

(b)$H{C_2}O_4^ - \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + {C_2}O_4^{2 - }$

(c)${H_2}{C_2}O_4^{} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2{H^ + } + {C_2}O_4^{2 - }$

The relationship between${K_{{a_1}}}$,${K_{{a_2}}}$and${K_{{a_3}}}$$ is given as :

  1. A ${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $+$ ${K_{{a_2}}}$
  2. B ${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $-$ ${K_{{a_2}}}$
  3. C ${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $/$ ${K_{{a_2}}}$
  4. D ${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $\times$ ${K_{{a_2}}}$ Correct answer

Solution

$$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \quad \mathrm{K}_{\mathrm{a}_{3}}$$ <br/><br/> $$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \quad \mathrm{K}_{\mathrm{a}_{1}}$$ <br/><br/> $$\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \quad \mathrm{K}_{\mathrm{a}_{2}}$$ <br/><br/> $$\mathrm{K}_{\mathrm{a}_{3}}=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right]}$$ <br/><br/> $$\mathrm{K}_{\mathrm{a}_{1}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right]}, \mathrm{K}_{\mathrm{a}_{2}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{-}\right]}{\left[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right]}$$ <br/><br/> $$\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}$$

About this question

Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp

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