Question

Data Part 1 Temperature of Ca(OH)2 solution: 20C Molarity of the HCl solution used for the titration: 0.05 M HCl Titration 1 Titration 2 Titration 3 Initial buret reading 10mL | 19 mi 2 mL 11 mL 28 mL 28 mm 36.4 q mL 8.4 ml Volume HCI used avg: 8.8 mL Part II Temperature of Ca(OH)2 solution: 84°C Molarity of the HCl solution used for the titration: 0:05 Titration 1 Titration 2 M HC Titration 3 Initial buret reading I 10 mL 15.5m. 22 mi Final buret reading 15.5 mL 21.4 m2 27.7 ml Volume HCl used 5.5 ml 5.9 mL 5.7 ml ava:5.7 ml Thermodynamics

Answer 1

Please refer below image that will help you;

Part 1 At 200 201 273= 2931 molarity of Hol = 0.05mHOL Averago volume HC required in titration - 8.8 mi Initial volume CaCOH)2 = lo mi we have roaction beth cacom 2 & HC OS, cacon & 2401 2 cache + 2H2O 0.01 8.8 ml HCl X o os molestici Imolcaconlax 2 moton 1000 m1 HER ^ 2 moine mol caebuia = 0.00049 moles on = 0.00044 -0.00am volume romi 1000 0010 we have relation of ksp and cons Ksp 2 con Room Temp Ksp= { 00497² = 8.518x10 4.259x10's AGE-RT enksp= -0.0821 x 293 en c4.259810s) Room Temp 06 = 105/188 = 242.08] part 7 To 84 C+273 Average volume HCL neqwned - 5.7 M = 357k 5.7 MIHCL X o.osmoles x 1 mol cacomer a molon 1000 mina 2 molno molca con on] = 0.000285 0.0285 m Ksp = - cont3 & (0.0285)3 = 1.157810S 0.01 TOMP AG:- RT en KSP -0.08217 357xen (H157x10°) High Temp AG = 333.166 J