You have an aqueous solution m
You have an aqueous solution made so that the formalconcentration of the soluble salt Na2SO3 is 0.15 M. Ka1 and Ka2 forH2SO3 are 1.2×10^-2 and 6.6×10^-8, respectively.
a.Write chemical equations showing all of the equilibria thatoccur in the solution.
b.Write the charge balance equation for the system.
c.Write a mass balance equation that relates the sulfurous andsodium species.
Answer:
a. Write chemical equations showing all of the equilibria thatoccur in the solution.
Water dissociates into H+ ion and OH–ion.
H2O (l) —–> H+ + OH–
The sulfite anion will act as a base and react with water toform the bisulfate ion, or
SO32-(aq) +H2O—–>HSO31-(aq) + OH–(aq)
The bisulfate ion can also act as a base and react with water toform sulfurous acid,
HSO31-(aq) + H2O(l) —->H2SO3(aq) +OH–
Sulfurous acid is doesn’t exist in aqueous solution. Itdissociates into sulfur dioxide,
When the bisulfate ion reacts with water, it’ll form
HSO31-(aq) + H2O(l) —–>SO2(aq) + H2O(l) +OH–
b. Write the charge balance equation for the system.
The concentrations of all the (+)’ly charged ions = theconcentrations of all the (-)’ly charged ions.
H2O (l) —–> H+ + OH–
[H+] +[Na+] –>2[SO32-]+ [OH–]
c. Write a mass balance equation
Sodium sulfate will dissociate completely in aqueous solution togive sodium cations and sulfite anions
Na2SO3 ——-> 2Na+(aq) +SO32-(aq)
1 mol of sodium sulfite will produce 2 moles of sodium cationsand 1 mole of sulfite anions.
[Na2SO3] = 0.15 M (given)
[Na+]=2*[Na2SO3] = 0.15 x 2=0.30M
[SO32-]= [Na2SO3] =0.15 M
Mass balance between Na+ andSO32-
[Na+] =2[SO32-]
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