Question

23°C Experiment 1 (Room Temperature): Temp. of Saturated Borax Solution: HCI Concentration: — 100M Trial 1 Volume of borax solution titrated: 5.00 ml Trial 2 5.00 ml Initial Burette Reading 3.20 ml 17.55 ml Final Burette Reading 17.55 ml 31.70 ml Volume of HCl used Moles of HCl used Moles of B.Os(OH), present in titrant Concentration of B.Os(OH), in titrant Concentration of Nain titrant Average Conc. of B.Os(OH)4 in titrant Average Conc. of Nat in titrant Equilibrium Constant Ksp expression Kp = Calculated equilibrium constant value Kop= Experiment 2 (Cooled Solution): Temp. of Saturated Borax Solution: HCI Concentration: 100M Trial 1 Volume of borax solution titrated: 5.00 ml Initial Burette Reading 15.00 ml Final Burette Reading 21.00 ml 4°C Trial 2 5.00 ml 21.00 ml 27.30 ml Volume of HCl used Moles of HCl used Moles of B.Os(OH) - present in titrant Concentration of B.Os(OH)in titrant Concentration of Nat in titrant Average Conc. of B.Os(OH), in titrant Average Conc. of Nat in titrant Equilibrium Constant Ksp expression Kop= Calculated equilibrium constant value (Continued on Reverse Side) TA Sig.

Answer 1

Earerimenti volume of He used = Final burette reading Initial burette reading = 17.55 mi - 3.20 ml = 14.35 mi molex of Ha ured = 14.35 mix conc. of Hol = 14.35 mix oum = 0.001435 molen Present = moles of HU used moles of Ba Og COH)2

= 0.001435 molel 2 = 0.00071ns moled 1. conc. of Baos Coule = 0.0007175 moles volume of borax solution - 0.0007175 molex smi = 0.1435 m conc. of wat = conc. of Beog (oh)x 2 = 0.1435 M X2 = 0.287m Trial 2 He used = 31.7-17.55 = 14.15mi moled of HCl = 14-15 mixo.im = 0.001415 moled moles of Btos (oh) = 0.001415 - 0.0007075

conc. of BAOS(OH)2 = 0.0007075 m I mole - Smi 0.1415m conc. of Nat = 0.1415m x 2 = 0.283M Average Nat come. = 0.283 +0.287 0.285 m Average B4 Os CoH)& conc = 0.1415+01435 5 = 0.1425 m Ksp = [at]CB495 Co4}}] Ksp = (0.285)? (0-1425) = 0.0115746

Trial 2 6.30mL Exra Trial I volume of He used .. 6 mL moles of Ha 0.0006 moled of Ba Os (GH) 0.0003 0.00063 0.000315 0.063 0:126 M Conc. of Baos come 0.06M conc. of wat 0.12m Average B495(OH). comc 0.0615 m. Average Nat conc. 0.123 m isp = (Nat]' [ 6405 (OH) -3 Ksp = 6-123) (0.0615) = 0.00093