Question

Calculate the concentrations of all species in a 1.07 M Na_2SCO_3 (sodium sulfite) solution. The ionization constants for sulfurous acid are K_a1 = 1.4 Times 10^-2 and K_a2 = 6.3 Times 10^-8. [Na^+] = [SO^2-_3] = [HSO^-_3] = [H_2SO_3] = [OH^-] = [H^+] =

Answer 1

Concentration of sodium sulfite (Na2SO3)-1.07 M Ionizati on of Na,SO, Na SO, (aą) O 2Na (a) SO3 (aa) Since Imol of Na,SO, forms 1mol of so3. [so] -[Na,so,]-107 M Since 1m ol of Na,SO, forms 2moles of Na', Na2[Na,so,]- 2(1.07M) = 214 M Equation representing the ionization of SO3 is as foll ows 2nd acidioni zati on constant of H2SO3, K -6.3x10 SO3isthe conjugate acid of base of HSO Baseionizati on constant of SO ,K^ - - ICEtable K 10x10* 16x10 K26.3x10-8 2- Initial concentrati on (M): 1.07 Change (M) Equilibrium concentration (M): 1.07-x LHso1 Baseionizati on constant, Kh - SO 1.6x10-7 107-х x=4. 1x10-4 M According tothe equilibrium table, x - oH- 4.1x10M According to the equilibrium table, HsGJ = 4.1 x 10-M According tothe equilibrium table. [so-1.07-x)M (1.07-4.110-)M 1.07M Equation representing the ionization of HSO is as follows -2 1st acidionization constant of H2SO3, Ka,-1.4x10 HS03 is the conjugate acid of base of H2SO3 Bas e ionization constant of HSO.K, = K1.0x1014 K,214x10 7.1x10-B

CEtable Initi al concentration (M): 4.1x10 Change (M) Equilibrium concentration (M) : 4.1x10 4.1x10 4.1x10 Baseionizati on constant, K^- HSO 7.1x10-13-(x)(4.1x10'+x) 4.1x10-4-x We can assumethat 4.1x10xM4.1x10M _(x)(41x10) 7.1x 10 141 x=7.1x10-13 M According to the equilibrium table, x [OH]- (4.1x10*+)M = (4.1x10+7.1x10-13)M -4.1x10+M According tothe equilibrium table, [H SO,4.1x10 M According tothe equilibrium table, [Hso]-(4.1x10*-x)M (4.1x10+-7.1x10 9)M -4.1x10+M 1.0x 10-4 4.1x 104 - 2.4x10-1M According to the ionic product of water, H - pHlgHg(24x0-") - 10.61