Find the concentrations of all ions in the solution at equilibrium after 0.213 L of 0.433 M aqueous Cu(NO3)2 solution is mixed with 0.213 L of 0.120 M KOH solution. Ksp for Cu(OH)2 is 1.1E-15.
[Cu][OH]^2 = Ksp .. ?
The Attempt at a Solution
ok so for the solution what i did was write out the balanced equation
Cu(NO3)2 + 2KOH -> Cu(OH)2 + 2K(NO3)
So we are told that the Cu(OH)2 = Ksp, so that means the spectator ions NO3 and K are going to be, their moles over the new volume. So i solved for them already and had received 0.061 mol/l K 0.433 mol NO3.
For Cu2+, what i did was i took the moles of Cu2+ initially, then i found moles KOH in terms of CuNO3, and i subtracted them, then divded the moles by the total volume to get 0.190 which is the same answer they got.
Now for OH, would i do the same process as above? Where i take the original moles of OH and then subtract it from moles of CuNO3 in terms of KOH, then divide that moles by the total volume?