Question

A solution of 50.0 mL of 0.045 M Ca(NO,), is prepared. What volume (in mL) of 1.90 M NaOH must be added to the calcium nitrate solution to begin to precipitate solid Ca(OH), from the solution? K = 6.5 x 10 Hint: Solid Ca(OH), will begin to precipitate at the point where an equilibrium exists between the solid and aqueous ions. The volume of NaOH required, will be very small so you don't have to consider the dilution of the AgNO, solution that occurs when the NaOH is added. Please note that you are adding NaOH to create a solution at equilibrium. Once prepared, no reaction takes place. Ca(OH),(s) Cal(aq) + 20H(aq) 0.0123 is entered 123E-2 in sci notation

Answer 1

Let's say ImL of 1.90(m) NaOH was added. so Total volume of the Solution = 50tv mL Now 50 mL 0.045M Ca(NO₃) = 0.045x50 milimote calNos), T 1000ml im CaNO₃ solution Contain I mote caNO₃)₂] 1 MLIM 1 " 10" s 1 milimole calNoz) e Similarly, V mL 1.90M Nash E 1.90v milimste NOOH Ca(NO₃), - cart & 2 Non NagH - Na+ + oH Ca(NO3) & NaoH both have good ionisation character. so No of milimotes of caht ins produced from card 2 = 0.045*50 No of milimoles of H ions produced from NaoH = 1.9OX V 0.045850 . [cat] = (M) 50 X V 0.3163 [OH-] = m2 1.90m 50 tv Nach should be CalGH) - ca 2+ + 204 added to start calot will start to precipitate when, 1.900V (m) the Precipitation of calott) [cat] [OH-] = ksp (CalotH), ) 0. 045x50 x (1.90v) = 6.5x10-6 50tv Isorn Now can we. or 0.045x50x3.61 vr = 6.5*10-6 (501) Vis very small compared to 50 (siven in hint) so assume, 50 tv x 50. 0.045x50x3.61 2 6.5x10-6 500x6.5x10-6 M, V 0.045X3.61 V 0.045*3.61 = 0.3163