Ka 1, Ka 2 and Ka 3 are the respective ionization constants for the following reactions (a), (b) and (c). (a) H 2 C 2 O 4 leftharpoons H ++ HC 2 O 4- (b) HC 2 O 4- leftharpoons H ++ HC 2 O 42- (c) H 2 C 2 O 4 leftharpoons 2 H ++ C 2 O 42- The relationship between K a 1, K a 2 and K a 3 is given as

Question

Ka 1, Ka 2 and Ka 3 are the respective ionization constants for the following reactions (a), (b) and (c). (a) H 2 C 2 O 4 leftharpoons H ++ HC 2 O 4- (b) HC 2 O 4- leftharpoons H ++ HC 2 O 42- (c) H 2 C 2 O 4 leftharpoons 2 H ++ C 2 O 42- The relationship between K a 1, K a 2 and K a 3 is given as (A) K a 3= K a 1+ K a 2 (B) K a 3= K a 1- K a 2 (C) K a 3= K a 1 / K a 2 (D) K a 3= K a 1 × K a 2

Tardigrade Admin · Accepted Answer

H 2 C 2 O 4 leftharpoons H ++ HC 2 O 4- K a 1 H 2 C 2 O 4- leftharpoons H ++ C 2 O 42- K a 2 H 2 C 2 O 4 leftharpoons 2 H ++ C 2 O 42- K a 3= K a 1 × K a 2

$K_{a_{3}} = K_{a_{1}} + K_{a_{2}}$

$K_{a_{3}} = K_{a_{1}} - K_{a_{2}}$

$K_{a_{3}} = K_{a_{1}} / K_{a_{2}}$

$K_{a_{3}} = K_{a_{1}} \times K_{a_{2}}$

$H_{2} C_{2} O_{4} ⇌ H^{+} + H C_{2} O_{4}^{-} K_{a_{1}}$
$H_{2} C_{2} O_{4}^{-} ⇌ H^{+} + C_{2} O_{4}^{2 -} K_{a_{2}}$
$H_{2} C_{2} O_{4} ⇌ 2 H^{+} + C_{2} O_{4}^{2 -} K_{a_{3}} = K_{a_{1}} \times K_{a_{2}}$

Ka3​​=Ka1​​+Ka2​​

Ka3​​=Ka1​​−Ka2​​

Ka3​​=Ka1​​/Ka2​​

Ka3​​=Ka1​​×Ka2​​

H2​C2​O4​⇌H++HC2​O4−​Ka1​​ H2​C2​O4−​⇌H++C2​O42−​Ka2​​ H2​C2​O4​⇌2H++C2​O42−​Ka3​​=Ka1​​×Ka2​​

$K_{a_{3}} = K_{a_{1}} + K_{a_{2}}$

$K_{a_{3}} = K_{a_{1}} - K_{a_{2}}$

$K_{a_{3}} = K_{a_{1}} / K_{a_{2}}$

$K_{a_{3}} = K_{a_{1}} \times K_{a_{2}}$

$H_{2} C_{2} O_{4} ⇌ H^{+} + H C_{2} O_{4}^{-} K_{a_{1}}$
$H_{2} C_{2} O_{4}^{-} ⇌ H^{+} + C_{2} O_{4}^{2 -} K_{a_{2}}$
$H_{2} C_{2} O_{4} ⇌ 2 H^{+} + C_{2} O_{4}^{2 -} K_{a_{3}} = K_{a_{1}} \times K_{a_{2}}$