Question

Solubility product constant of calcium iodate/common ion effect.

I am unsure how to do this however I am only asking for help on trial one as there are 4 total trials and I would like to be able to do those on my own after I have learned how to do the work. Thanks.

Also, could you provide the balanced equation for the titration as well as the solubility equation for iodate?

Part A: Saturated Calcium lodate-No Added Calcium lon Trial #1 Trial # 2 Concentration of Sodium Thiosulfate Solution (M) 0.0100M Volume of Caleium lodate Aliquot Added in Liters 0.0050し Initial Volume Thiosulfate Solution (mL)-from buret (Read buret directly-it is designed to be read without changing) Final Volume Thiosulfate Solution (mL) from buret SOAL 35.9mL Volume Thiosulfate Solution added (mL) 14.1mL 13.8ml (Final volume minus initial volume on buret) Total Volume Thiosulfate Solution added in liters (L) Moles Sodium Thiosulfate added (M,V,- #moles) Moles lodate present (using titration stoichiometric relationship) See calculations section for balanced chemical equation IO, in saturated solution Ca in saturated solution Molar Solubility of Calcium Iodate (M) Ksp value Calcium lodate Average Ksp value for Calcium lodate olubility product equation for calcium iodate (with constant value):

Answer 1

0 In this case [ 10,is determined by the standarisation with Sodicem thiosulfate solution is presence of KI Solution and Starch as indicatore. The reactions are 10 caq) + 5I®(@q) + 6Hgo (aq) -> 31262 +9620(1) :31, US +65205 (aq) → 619 +35406 = (aq) 10g (aq) + 65,03 (29) + 6H3O+ (aq) → Ilaq) +35206 (29) . 1 moles 67 103 = 6 moles of $202. som I +2 +010) Trial I I Talal-2 [Na2S203) solution 0.01 CMS volume of Ca(103), sol 0.0050 L Initical volume of NOOS 02 sota 50 ml • 50 ml Final volume of NG SOz solm 35.90 ml 36 • 20ml volume of NaSaoe som added 50-35-90) = 14.40 mb 50-36.20) - 13:80 mk. I volume of Na2S203 sohn added (L)_ 0.0141 0.0138 Moles of Na, S.O2 (MV) 10. 01X0.0141 = 1041X13 0:01X00138 = 1.38x10 | Moles of 103 present 11.41710/6) (1.39 *10916) = 2:35.410-5 = 2:30 2100 [103 ) = Moles of 103/volume) (2:35x1050.0050) 1(2130x105/01005) = 4,70x10° (M) = 41604103(M) [Ca2+] =[105] |(4.70*10*12) (A-G0X1 1-2-35 x 15 3 (M) 2:30 X 15 3 (M) Molar solubility of Ca (013)2 -2.35 x103 CM) - 2:30 × 103 (M) 'Ksp I ca(103)2] = [Ca2+] [102]-|(2-35x104.70*10') (2:30x20°) (4.6x103)? = 5.19 x 10 9 = 4.87x108 Average Ksp LCa(103)2] 6-19+4•87)x108 = 5:03ties 2 ... The answer is ksp [ca(103)2] = 5.03X10°