Question

03 Question (1 point) The solubility of slaked lime, Ca(OH)2, in water is 0.185 g/100.0 mL. 1st attempt Feedback IM! See What volume of 3.00x103 MHCl is needed to neutralize 13.5 mL of a saturated Ca(OH)2 solution? * 227 x 104 mL 2 OF 4 QUESTIONS COMPLETED 030 > + VIEW:

Answer 1

ca(OH)₂ +2H U cael + 2H₂O. so caloh) is an st nong base and . Au is a strong and . . so Hell added in calor), form en el so to neutralise are 1 mole of calor), we need 2 mol of Hel. Solubility of calow), = 0.1859/100.me so as per the definition of solubility. 00185 g of solite is the maximum amount that can be dissolved in ioo me of golvent so we can say find the conenfisation no when maximum solute is added. so concentration = 0.185 g. molar mass of coloH)2 g/mol x1 concentration - 0135g 0.185 mol - 74.092 g/mol x100 7409.2 m2 30 mol present in cabu), of volume 13.5 m/ i 0.185 7.1 3.5 mol 2409.2 " ; so nand more of Hel need to neutralize. arts 2* 0.185 X13:5 mot and Molamily of Hel (given) 2 3x10-3 mol 1000 ml so volume of Hell required 1000 m2 g 2 x 0:185 x13.5 mol 3x10-3 mol 9409.2 s po 2.247 X 10 -4 X106 m2 2 me 2 2:24zx10² m2 X 100 ml