posted by Dalala on .
lets say we have Cu(NO3)2 and then we add NaOH. so Cu(NO3)2 + NaOH and then we add HNO3 to it what wil happen?
then we take Cu(NO3)2 + NaOH + NaNO3 what will happen then?
and what would happen if the order was reversed for example Cu(NO3)2 + NaNO3 and later we added the NaOH
I like to know where we're going so I don't buy a pig-in-a-poke but on the basis of what you wrote.
Cu(NO3)2 + NaOH ==> Cu(OH)2(s) , that is, copper(II) hydroxide precipitates. Adding HNO3 just dissolves it again because the HNO3 neutralizes the NaOH, the solution is not basic anymore, and Cu(NO3)2 is soluble in neutral to acid solution. Addition of NaNO3 has no effect either way it is added. The bottom line is
Cu^+2 + 2OH^- --> Cu(OH)2(s). When the hydroxide ion is large enough, along with the Cu(II) ion, so that Ksp is exceeded for Cu(OH)2, a ppt will occur.
By the way, the cloudy phase you saw in the NaCl/water/acetone experiment, probably was NaCl coming out of aqueous solution since NaCl is very insoluble in acetone.