Let us assume that Cu(OH)2(s) is completely insoluble, which signifies that the precipitation reaction with NaOH(aq) (presented in the transition) would go to completion. If you had a 0.300 L solution containing 0.0230 M of Cu2+(aq), and you wished to add enough 1.31 M NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation. Express the volume to three significant figures and include the appropriate units.
To determine the minimum amount of NaOH(aq) solution needed to precipitate all the Cu2+(aq), we can use the stoichiometry of the reaction:
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(s)
From the balanced equation, we can see that 1 mole of Cu2+(aq) reacts with 2 moles of OH-(aq) to produce 1 mole of Cu(OH)2(s).
Given:
Volume of Cu2+(aq) solution = 0.300 L
Concentration of Cu2+(aq) = 0.0230 M
Concentration of NaOH(aq) = 1.31 M
First, calculate the moles of Cu2+(aq) in the solution:
moles of Cu2+(aq) = concentration x volume = 0.0230 M x 0.300 L = 0.0069 moles
As 1 mole of Cu2+(aq) reacts with 2 moles of OH-(aq), we need twice the amount of moles of OH- ions to precipitate all the Cu2+(aq).
Moles of OH-(aq) needed = 2 x moles of Cu2+(aq) = 2 x 0.0069 moles = 0.0138 moles
Now, let's calculate the volume of NaOH(aq) needed to provide the required moles of OH-(aq).
moles of NaOH(aq) = concentration x volume
0.0138 moles = 1.31 M x volume
volume = 0.0138 moles / 1.31 M ≈ 0.0105 L
The minimum amount of NaOH(aq) solution needed is approximately 0.0105 L.