Question

What ae the concentrations of HSO 3 . and SO 3 2- in a 5.45 M aqueous solution of H 250 3? Ka 1 = 1.2 x 10-2 and Ka 2 = 6.6 x 10-8 O [HSO3] = 0.342 M; [S032.1 - 1.2 x 10-2 M [HSO3) = 0.256 M; [SO32-1 - 2.6 x 10-7 M [HSO3) = 0.256 M; [S032-1 = 6.6 x 10-8 M O [HSO3) = 0.182 M; [SO32-) = 0.0032 M O [HSO3) = 2.6 x 10-7 M; [S0321 = 0.465 M

Answer 1

-2 Sol?=> what are the concentrations of HSO3 and So in a 5.45m lucous solution of H2SO3, Key = 1.2 x 10 & Keiz=668108 => correct option is (3) j.e. (H503] -0.256 m; Coukucation - Given! [ 50327 =656410-8m concentration of H2503 = 5.45 m ( sulfurous aid) Kai = 1.2x102 & Ka₂=6*6* 10 =) H2803 is a weeck acid of dissociate in . H2503 = H+ + HS03 I 5:45 -8 С. -e it ze te C 5.45-ze ze. e Kais e x ze THAJ [1503] [H2SO3 5.45-se 2 e2 1.2 x 10 = 5.45-20 5.45 = 454.16 1. 2x152 5.45 1.2XL02 1.2 x 102 22 .. > Loo 5:45 then re is very small f ::be=(.2x703x65-45) ze² = 0.0654 ::2= V0.0654 0.2557 0.256 m neglect it. -

Cht) + 2- H 503 H C 1.716 X 10^8 Ha H503] ::C1503] =0.256 m concentrihon of H503 = 0.256 m =) Calculate Concentration of su3. Dissociation of H50374 ICE table is HSO3 + 0.256 m 01256 - y ty- +7 0.256+ y - [H] [503 Kaza (0.256+Y.J X Y [HSog 0:256-Y 6.6* 10 (0.258+.Y) X Y 0.256-7 then y is very 6.67 10 0.2567 small f neglet it :.01256 Y =(E16X10 ) x (0.256) 1.7 16 X 10" Y = 6.6 x 10 Hence Comeet option Y = [503 6.6X16+'=> E 0.256-Y. Y. = 0.256 6.6x168 > 100 -8 = 0,256 -8 - 1000 2 0.256 is (3),