Question

Chem 1212 Lab Manual - Revised 06/2018 Molarity of HCI .05565 ) * Trial #1 Volume of Ca(OH)2 10.00 ml (1 p.) Initial Buret Reading 0.00ml Final Buret Reading 8.00ml Volume HCI Delivered 8.00ml Trial #2 230 1000mline 8. 30ml 16,20ml 7,90 ml _(4 pts.) Solubility (M) Kop of Ca(OH)2 _(4 pts.) (4 pts.) Sample calculation of solubility (M) of Ca(OH)2 in deionized H20, Trial 1. (4 pes.) Helhores x .05565 * Sample Calculation of Kp, Trial 1.4pts.) Average Solubility (M) (pt) Average Ks Average Solubility (g/100mL) pe) Average pkp Calculation of average solubility of Ca(OH), in g/100 mL. (4 pts.)

Answer 1

Here solution of calot must be saturated. At saturated condition the expression of solubility froduct 2 r. Calotte disse- Kop & Gear & Co op - ca Lciates as u..6 L Calotta (s) Caton) set, solubility of ca Cote is 5'r Cart - sof Cota 25m .: using rauation (i) we can write +2014109) kop - sx(25) = 48 kob - 453 .....) Now, Ho react with oth which is present at saturated solution of casolth as: Heat Oltar – Holt ela)

for tria 1 in an 8.oom of 0.05565 Hel the morser of mot of Hel - 0 05565 molt & zout - 4.452 x 10 mol . According to stoichiometric ratio of the reaction (ii) mot Her rennt with mot ot- 4.45280 & & & 4.452810 saturated de 10.0me of calotte solution Ott present 4.452 x 18 mot . Der saturated solution Cost = 4.452x15 mol wol x 180 - G. 452 x 10 2 wise - 0.04452 M. n 25 - 0.04452 in so s. 04452 - 0.0 2226 m solubility of casolte in deionised tho, Trials in 0.022 26th sample calculation of kofi kob- S4 (25) T & 453 & 4 x (0.02226) = 94&10-S in kopu 4.48105

Similarly for trials moles of Hel used = 0.05 565 molx x 7.900 = 4.896 *10-4 mos De 10 ml of coolt, soln oth freneur G.396x15 4 in i Colt- - - 4.3968104 mol 100 9 ) 12 - 0.04 396 in 25 = 0.04306 M. S - 0'09396 m = 002198 M kot = 4x (0.02198)3 (usong Cis) ) 425 X 10" = 4.364 i Average solubility - Lo10222 6+ 0.09198 m o'Onn m in Average kop & y [ 4.4x1025 + 425 4105 & 4.3215x10 "I # Averagep Kop & -log kuß – – log (4.32,5 x 10 s) V rub in Average prop= 4.364. : Average solubility 0.0221M . 1000 ml solution contain 0.02212 nol Carolt 0.022/2 7850 .: 70 a . 0.02212 AIRY .. Arg. solubility [mol man of 1088 caphe 741093ehmest 0.0024ranil 0.002 uplx 74093 = 0.1578/100m