Question

Calculate the concentrations of all species in a 1.61 M Na,SO2 (sodium sulfite) solution. The ionization constants for = 1.4 x 10 and K2 = 6.3 x 10-8. sulfurous acid are Kal Na* М HSO5 H,SO,] М [он ] - H М M M M

Answer 1

Q. Given that molarity of sodium sulfite (da2503) Solution - 1.61m The ionization constants for a - acid are Ka,= 1.4x102 Kaz=6.3x108 we know that acid dissociation constant (ka)x base dissociation haton constant (kn-kwalxlok constanta le. KaXK6 = kw=1x1014 - ③ We know that sodium sulkite solution (da2503) dissociates into cons in aqueous solution. Narsog (aa) 2 Nat (aq) + sor (aq) Here, Nagsoz contain twolz) dat ions. Concentration of dat = [ dat] - number of dat cons x enolarity = 2x1.6IM = 3.22M Hence, Inat] =3.22M [soz JZ1X1.61m = 1.61M [soz -] Now 1 Kaq 6.3X100 Base dissociation constant (Kb) for Soz - = Kw - lxloka6 (from equation Kbs = 11.58x107 Base dissociation constant (kb 2l for soze = kw - 1x1014 - Kas - 1.4x102= 7.1x1013 Kbz for sos = 17.1x1013)

Now, first dissociation of soz- soz- (aq) + H2O(4) Hoz (aal tot (aa). initial 1.61M Concentration change -XM 4 хтм Хм At equilibrium (1.61-XM Xм Хм Now, suppose that 1. ьі м>>> Хм So, (1.61-x)m a 1.61M first base dissociation constant (K6L) = [Hsoj ] [ori} I 1.588 157 [soz- ] (x) (x) me - - 1.58x107 1.61M x² = 1:58x10-7x1.61 m. x = 1 1.58x157x1.61 M = 5.04x104m [Hsong ] = [or] =5.04x154 m) dow. Second dissociation of Hsuzlaq) + H2000 Hz soglaa) + oH BU initial 5.04x104m om souxiom concentration Am +AM TAM change in concentration At equilibrium (5.04xl0 AM AM 6:04x10tall second base dissociation constant(kbz! 12.503] [ori - 7.1x1013 [Hoj]

Assume that - soux1044 AM s So [hsor] =(5.04x104_AM =5.04x10 4M Tor] =(5.04x10"+AM = 5:04x100 m dow, K bz = A x SouX10 4 MR = 7.1x1013 A = 7.10. 1613 M . [H2503] = TA] = 71816Bm sulfurous acid we know that kw.= [H] [ok] = lxlokk TH+] - IX1014 - lxlo14 [ot ] - sioux104 [H+] = 1.98 xion Hence i Edat] =3.22 M, [so}}=1.61M [Hsoj] = 5.04x104m [H₂.5033 = 7.181013 TOH J = 5.04x104m [H+ ] = 1.98 81611M