what will be the observations for the reaction:
CuSO4(aq) + NaOH(aq)
its a two mark question so two observations at least.

Question

what will be the observations for the reaction:
CuSO4(aq) + NaOH(aq) 
its a two mark question so two observations at least.

thomaster · Answer

$$CuSO_{4} (aq)+ NaOH(aq) \rightarrow CuOH (s) + NaSO_{4}(aq)$$
You see the reaction produces a solid so one observation will be the precipitation of Copper(I)Hydroxide (which will be a blue/purple precipitate).

anonymous · Answer

hmmm... sodium (or any alkali metal) cannot carry a two-plus charge. well, it won't be easy because we are stripping the second electron from the neon core. 

the way you've written it suggests that the Cu^2+ somehow oxidizes the Na+ cation. that takes too much energy to be observed.

anonymous · Answer

i am going to guess we will observe Cu(II)(OH)2 instead. it has low solubility so it will show up as a precipitate. the other two ions, sodium and sulfate are spectators.

thomaster · Answer

lol i see i made a mistake, it should be Na2SO4
So the reaction will be:
$$CuSO_{4}(aq) + 2NaOH(aq) \rightarrow  Cu(OH)_{2}(s) + Na_{2}SO_{4}(aq)$$

thomaster · Answer

copper(II)hydroxide will be a blue precipitate