Question

MOC0sn 8. Will KCI04 precipitate when 20.0 mL of a 0.050 M K* solution is added to 80.0 mL of a 0.50 M CIO4 solution? Mstdo muis olidullont oo Ksp for KCIO4 is 1.05 x 102

Answer 1

Solution is added to 80.0 ml 2 20.0mL of a 0.050 M K+ of a 0.750 M CLOT Solution. Ksp of Kuloy KCI04 (8) = x+(aq) + Ciot (aq) Ksp = [kt] [ciou] ionie.product of = 1.05*10-2 kt/ & [co ] = [ktda. product is equal to the Ksp or greater than solid Kelou being precipitated. If the asp the ionic the - 20. me of 0.050 M at Solution is added to 80.0 me of a 0.50 M UOL Solution Se, Volume of total Solution = 100.0mL Now, 0.050 M of 200ml of *+ Solution = 0.050 ) milimol of kt 2 1 mili mol of kt present.

And 80.0 ml of 0.50 Mof drou - (30.0x0.50) emilia mili mol aby present = 40 milimol. I milimol of xt will be present in 100 me of solution And 40 milimel of clou will present also. loome Now, 1 milimole of at will be present at of Solutim. 1x 10-3 (mol) of Kto - So, concentration Isomi Xio = 10-2 (M) concentration of cloy = - 40x105xloun (m) Similarly - 400 x10-3 My = 0.4M) contion at So, Now, ionic product of the solution and croy = (0.4 x 10-2] - 4X10-3 < Ksp So, Ionic product is less than ksp so, No precipitate will observe.