Question

Sulfurous acid (H2SO3) is a diprotic acid, (Ka1 - 17 10-2 and Ka2 - 6,0 x 10-8 what is the equilibrium concentration of SO3? if the initial concentration of H2SO3 is 1.2 M? O 6.0 x 10-8 M O2.0 x 10-2 M O 1.7 x 10-8 M O 17 x 10-2M 7.2 x 10-8

Answer 1

SOLUTION H₂SO3 is a diprotic acid. brivent ka - 1.7 x ka - 6.0 x 10-8 [H₂SO3] 1.2M 1st ionization : H₂SO3 Lag) Hit Laq) + 450 Initial 1.2M Equilibrium 1o2on (og) i. Ka, - [t] [HSO₃] [H₂SO3] - Kai as n is small 102-2 - 2.|| ء ) - O: V 1.7x10-2x 1.2 0.143 [ht]-[H502] = 0.143 and ionization in 4150₃ - Lag - Ht & So, 2- Initial 00143 00143 0 Equilibrium D-143-4 0.143ty ty un ka - [47] [S0₂ 2] [HS03]

Sence and yer 0.143 so, 0.143 ty 7 0.143 0.143-y – 0.143 ka= [0-143 ty] [S03²-] (0.143-y) ka, = 0.143 *[S03 ² ] 0143 [so 2] = kag - 6.0 x 10-8 a M. 08 27 give The equation is a NHL I LES & NH₂ (9) & HI Lg). We know that, kp - IN 1kp = PNH PHI - where PNHg and Pri represent partial pressure of NH₃ and Hi at equilibrium. NOTE & Initially only why I ls) was introduced in containes so at equilibrium, IPNH₂ = PHI 16 Using relations and D atri kp- (Pong 2 u. Pars & ke 10. 215 atm atm - 0.464 at equilibrium un partial is of ammonia atm. pressure S. 464

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