162 Part B: Calcium lodate with 00500 M Calcium lons Trial #1 Trial #2 Volume of

Question

162 Part B: Calcium lodate with 00500 M Calcium lons Trial #1 Trial #2 Volume of Calcium lodate with Calcium Nitrate Solution Aliquot Added (L) | 5 m L-???st Initial Volume Thiosulfate Solution (mL)-from buret oOmL Final Volume Thiosulfate Solution (mL) from buret a7.9 m 7.5ml Total Volume Thiosulfate Solution (mL) (Final minus initial Total Volume Thiosulfate Solution (L) 47.9 mu 67. Sml Moles Sodium Thiosulfate added moles muti 1 Moles lodate present (using titration stoichiometric relationship) see balanced chemical equation background 0, 0 002 325 .00029 mo les IIOs in saturated solution Predicted |IO,1 with 0.0500M Ca2t ions See background for calculations Ca lcstos from Calcium lodate dissolving See background section Average Molar Solubility of Cale Added

Explanation / Answer

IO3-(aq) +6S2O32- (aq)+6H+(aq)3S4O62-(aq)+I- (aq) +3H2O(l)

So,IO3-(aq) reacts with S2O32- (aq) in 1:6 molar ratio.

[IO3-]/[S2O32-]=1/6

6[IO3-]=[S2O32-]

Using Equation,

V(thio)*M(thio)=6*V(iodate)*M(iodate)

where,V(thio)=volume of thiosulfate

M(thio)=molarity of thio

V(iodate)=volumeof iodate

M(iodate)=molarity of iodate

trial 1) mol of thiosulfate reacted=0.05M*0.0079L=0.000395 mol=V(thio)*M(thio)

So mol of iodate reacted=V(iodate)*M(iodate)=mol of thiosulfate reacted/6=0.000395 /6=0.0000658 mol

[IO3-]=mol of iodate/total volume=0.000658 mol/(5ml+7.9ml)=0.0000658 mol/0.0129L=0.0051 M

predictd [IO3-] with 0.05M Ca2+ =0.0051M

Ca(IO3)2 <--->Ca2+(aq) +2IO3-(aq)

[Ca2+]CaIO3 from calcium iodate dissolving=1/2[IO3-]=1/2*0.0051M=0.00255M

trial 2

mol of thiosulfate reacted=0.05M*0.0075L=0.000375 mol=V(thio)*M(thio)

So mol of iodate reacted=V(iodate)*M(iodate)=mol of thiosulfate reacted/6=0.000375 /6=0.0000625 mol

[IO3-]=mol of iodate/total volume=0.0000658 mol/(5ml+7.5ml)=0.0000625 mol/0.0125L=0.005 M

predictd [IO3-] with 0.05M Ca2+ =0.005M

Ca(IO3)2 <--->Ca2+(aq) +2IO3-(aq)

[Ca2+]CaIO3 from calcium iodate dissolving=1/2[IO3-]=1/2*0.005M=0.0025M

Average molar solubility of calcium iodate with calcium nitrate added=(0.00251+0.0.250)/2=0.002505M

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