![Using the balanced equations given below, determine the overall molar ratio between aqueous
oxygen and thiosulfate. Using the known concentration and added volume of sodium thiosulfate
solution along with this molar ratio, it is possible to calculate the amount and concentration of
dissolved oxygen in the original sample. Determination of Dissolved Oxygen. Our determination
of dissolved oxygen relies on several coupled reactions. Ultimately, the oxidation of manganese
by dissolved oxygen drives oxidation of iodide to iodine, which is titrated using the iodine-
reducing agent sodium thiosulfate. The water sample is first treated with a manganese(II)
solution and a solution of potassium Mn2+(aq) + 2 OH-(aq) Mn(OH)2(s) oxygen dissolved in
the water oxidizes a portion of the manganese( hydroxide to manganese(ID hydroxide. (Note that
dissolved oxygen is the limiting reagent throughout these reactions.) To dissolve the
manganese(III) cations back in the water completely, a small amount of sulfuric acid is added--
the acid protonates the hydroxides, allowing the freed manganese(I) ions to return to solution. 4
Mn(OH)2(s)-O2(aq) + 2 H20() 4 Min(OH)3(s) 2 Mn(OH)3(s) + 3 HSO4(aq) 2 Min'"(aq) + 3
SO42-(ag) + 6 H20() Iodide now enters the picture, as the newly dissolved manganese(II) ions
have the ability to oxidize iodide. Dissolved molecular iodine forms, producing a yellow-brown
solution. Iodide remaining in solution coordinates to the newly produced iodine, yielding the
triiodide ion which turns blue in the presence of dissolved starch. L(aq) + rag) 13-(aq) Finally,
sodium thiosulfate solution is added dropwise to the mixture until all of the triiodide has been
reduced back to iodide. At this point, the blue color of the starch-triiodide solution will
disappear, leaving a colorless solution.
Solution
The mole ratio between the related species in the series of rxns are as follows:
[I3-]/[S2O32-]=1:2 (mol ratio)
[I2]/[I3-]=1:1
[Mn3+]/[I2]=2:1
[Mn(OH)3]/[Mn3+]=2:2
[O2]/[Mn(OH)3]=1:4
multiply all the equation,
[I3-]/[S2O32-]*([I2]/[I3-])*([Mn3+]/[I2])*( [Mn(OH)3]/[Mn3+])*
([O2]/[Mn(OH)3])=(1/2)*(1/1)*(2/1)*(2/2)*(1/4)=2/8=1/4
[O2]/[S2O3]=1:4 mole ratio
Using the known concentration(C(S2O3) and added volume (V(S2O3)) of sodium thiosulfate
solution](https://image.slidesharecdn.com/usingthebalancedequationsgivenbelowdeterminetheoverallmola-230707211352-538a8e78/85/Using-the-balanced-equations-given-below-determine-the-overall-mola-pdf-1-320.jpg)
![[S2O3]mol=(C(S2O3) *(V(S2O3))
So,[O2]mol=1/4*[S2O3]mol=1/4*(C(S2O3) *(V(S2O3))
[O2]mol=V(O2)*C(O2)=1/4*(C(S2O3) *(V(S2O3)) ......#
C(O2)=1/4*(C(S2O3) *(V(S2O3))/V(O2)
So,the amount (mol) of dissolved oxygen can be determined using # ,but to determine the
concentration ,volume of O2 =volume of sample is required](https://image.slidesharecdn.com/usingthebalancedequationsgivenbelowdeterminetheoverallmola-230707211352-538a8e78/85/Using-the-balanced-equations-given-below-determine-the-overall-mola-pdf-2-320.jpg)
Using the balanced equations given below, determine the overall mola.pdf
- 1. Using the balanced equations given below, determine the overall molar ratio between aqueous oxygen and thiosulfate. Using the known concentration and added volume of sodium thiosulfate solution along with this molar ratio, it is possible to calculate the amount and concentration of dissolved oxygen in the original sample. Determination of Dissolved Oxygen. Our determination of dissolved oxygen relies on several coupled reactions. Ultimately, the oxidation of manganese by dissolved oxygen drives oxidation of iodide to iodine, which is titrated using the iodine- reducing agent sodium thiosulfate. The water sample is first treated with a manganese(II) solution and a solution of potassium Mn2+(aq) + 2 OH-(aq) Mn(OH)2(s) oxygen dissolved in the water oxidizes a portion of the manganese( hydroxide to manganese(ID hydroxide. (Note that dissolved oxygen is the limiting reagent throughout these reactions.) To dissolve the manganese(III) cations back in the water completely, a small amount of sulfuric acid is added-- the acid protonates the hydroxides, allowing the freed manganese(I) ions to return to solution. 4 Mn(OH)2(s)-O2(aq) + 2 H20() 4 Min(OH)3(s) 2 Mn(OH)3(s) + 3 HSO4(aq) 2 Min'"(aq) + 3 SO42-(ag) + 6 H20() Iodide now enters the picture, as the newly dissolved manganese(II) ions have the ability to oxidize iodide. Dissolved molecular iodine forms, producing a yellow-brown solution. Iodide remaining in solution coordinates to the newly produced iodine, yielding the triiodide ion which turns blue in the presence of dissolved starch. L(aq) + rag) 13-(aq) Finally, sodium thiosulfate solution is added dropwise to the mixture until all of the triiodide has been reduced back to iodide. At this point, the blue color of the starch-triiodide solution will disappear, leaving a colorless solution. Solution The mole ratio between the related species in the series of rxns are as follows: [I3-]/[S2O32-]=1:2 (mol ratio) [I2]/[I3-]=1:1 [Mn3+]/[I2]=2:1 [Mn(OH)3]/[Mn3+]=2:2 [O2]/[Mn(OH)3]=1:4 multiply all the equation, [I3-]/[S2O32-]*([I2]/[I3-])*([Mn3+]/[I2])*( [Mn(OH)3]/[Mn3+])* ([O2]/[Mn(OH)3])=(1/2)*(1/1)*(2/1)*(2/2)*(1/4)=2/8=1/4 [O2]/[S2O3]=1:4 mole ratio Using the known concentration(C(S2O3) and added volume (V(S2O3)) of sodium thiosulfate solution
- 2. [S2O3]mol=(C(S2O3) *(V(S2O3)) So,[O2]mol=1/4*[S2O3]mol=1/4*(C(S2O3) *(V(S2O3)) [O2]mol=V(O2)*C(O2)=1/4*(C(S2O3) *(V(S2O3)) ......# C(O2)=1/4*(C(S2O3) *(V(S2O3))/V(O2) So,the amount (mol) of dissolved oxygen can be determined using # ,but to determine the concentration ,volume of O2 =volume of sample is required