Question

(10pts) Caloulate the molarty of each hpe of ion present in 0.00202 M AbSO solution 3 4 (10pts) What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution? 5. (10pts) How many miliners of a 10.15 M NaOH stock solution are needed to prepare15.0 L of 0.315 M NaOH? 6. (10pts) What volume, in ml, of 0.40 M HCI neutralizes 29 29 ml of 0.17 Calcium Hydroxide? M

Answer 1

* Given metanfty (M) g Algo), solution = 0.0000am solution: molarity 9 each sons can be calculated as Al(sly), - 2443+ + 350,45 Pottally there are 2 Alions & 3 sulfatt ions are present. .. melantity 9 A13+ Yons = 2X0.00.202 M = 0.00404 M niy malarity of soor ons = 3x 0.0D 202 M 0.00606 M

e given : 33378 potassium hydougen cwbonate (KH (0) solutiu: molarity - n g meles Volume y solution in mass / molar mass 3338 100.1%g mot - = 0.333 moll- 10.0L 1 molarity of KHCO = 0.333 mollt = 0.333 m) © wikit M,V, = mv, M, my -melarities > V, - m, va Nivo - volume o solution mi - 15.02 x 0.315M - 0.466L @ 466 milli litre 10.15M - required Radh - 466.0 millflitoes = 466 ml = 0.486L) © Reaction bloo Hel and calohcan be written as 2 Hd + Ca(OH)2 - Cachy + 2HO .. a metes g Hel 94uivis one mole of Ca (OH), to pooduce product. meles - volume x concentration - 29.29X0.17 = 4.98 moles, calm, i moles of Hill requires 4.98 moles of calolHz => x= 4.98 x2 = 9.96 moles og Hel. : volume of Hell required to molls - 9.96 - 24.90 mi concentration 0.40 necetralize T volume 9 Het required - 24990 m/s 24.90 miltelike