Question

1. (3 points) Calculate the equilibrium concentrations of the different species in a 0.0100 M solution of sulfurous acid, H₂SO₃, at pH = 4.00.

Answer 1

H2SO3 → H+ +H 50 Ka, = 1.4x162 HSOğ 2 H+ + 507 ka = 6.3x108 H₂SO₃ > с-сха H+ + HSO3 ((x + (4,92) (a) ki - e-cal +taya, ca, kz см, – H+ + 50% = (x + Capa ) Card ca - Ca&2 Cx + Cap de Cra2 cx, x2 HSO3 Cx, - CAP2 [ht] =CA, + (0,2 (2nd equilibrium ca L constant is very low compared to first one pH = 4 de X, 742 .: Ca, + (x, & ald : - log ex, = 4 - log o.ola, = 4 0.01% = 104 x = 10

kzCX, 22 in 6.3x108 = 0.01 x 102 x d. de = 6.3X164 i. [14505 Eso Ca = 104 ] =(x,- (x, a ~ cx = 10 [H₂SO₃ = b c-ca, = 0.01 – 0.01 102 = 0.99x10? ~ 0.01 K2 = [ht] [50²] [HS01 K2 [H0] en [so27 = [H+2 = 6:3x10 x 100 = 6.3x10 - [S03") = 6:3x108

[H+] = 104 meillit [HS03 ) = 104 mol / lit [H2 503 ] = 10² mollit [s02-] = 6.3x108 mollit