Question

The following drawings represent aqueous solutions. Solution A is 2.00 L of a 2.00M aqueous solution of copper (II) nitrate. Solution B is 2.00 L of a 3.00 M aqueous solution of potassium hydroxide.

a. Draw a picture of the solution made by mixing solutions A and B together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared with solutions A and B and the correct relative number of ions, along with the correct relative amount of solid formed.
b. Determine the concentrations (in M) of all ions left in solution (from part a) and the mass of solid formed.

Short answer

a) The picture of the solution made by mixing solutions A and B together after the precipitation reaction takes place.

b) The concentrations (in M) of all ions left in solution is$\left[C{u}^{2+}\right]=0.250mol$and the mass of solid formed is $Cu{\left(OH\right)}_2$is$292.65g$

Step-by-step solution

Definition 

Themolar mass of a material is defined asthe mass of the substance contained inone mole of the substance, and it isequal to the total mass of the component atoms. It may be stated mathematically as follows:
Molar mass$=\frac{Weighting}{numberofmoles}$

Determination of chemical reaction

a)

Copper hydroxide is precipitated when copper nitrate is combined with potassium hydroxide.

$C{u}^{2+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)+2K^{+}+2O{H}^{-}\to Cu{\left(OH\right)}_2\left(s\right)\downarrow +2K^{+}+2N{O}_3^{-}\left(aq\right)$

Determination of net precipitation reaction 

b)

Moles of$Cu{\left(N{O}_3\right)}_2=\left(Molarity\right)\left(VolumeinL\right)$

$=\left(2.00M\right)\left(2.00L\right)$

$=4.00moles$

$\therefore \left[C{u}^{2+}\right]=4.00moles$

$\left[N{O}_3^{-}\right]=2\left(4.00moles\right)$

$=8.00moles$

Molesof$KOH=\left(3.00M\right)\left(2.00L\right)$

$=6.00moles$

$\therefore \left[K^T\right]=6.00moles$

$\left[O{H}^{-}\right]=6.00moles$

Therefore the Net precipitation reaction is,

$C{u}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\to Cu{\left(OH\right)}_2\left(s\right)\downarrow$

Determination of total Volume of the solution mixture

Here 2 moles of$O{H}^{-}$ will react with 1 mole of$C{u}^{2+}$ to create 1 mole of $Cu{\left(OH\right)}_2$precipitate.

Thus, $6.00moles$of$O{H}^{-}$will react with $\frac{6.00moles}{2}=3.00moles$of$C{u}^{2+}$ to produce $3.00moles$$Cu{\left(OH\right)}_2$precipitate.

Let’s compute,

Molar massof$Cu{\left(OH\right)}_2=63.55+2(16+1)$

$=97.55\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Therefore the mass of formed solid$Cu{\left(OH\right)}_2=\left(moles\right)\left(molarmass\right)$

$=\left(3.00\right)\left(97.55\right)$

$=292.65g$

Therefore the ions left in the given solution,

So, total volume $=2.00L+2.00L$

$=4.00L$

Determination of concentration of [Cu2+] 

Let’s compute theconcentration,

Concentration$=\frac{Moles}{Volume\left(inL\right)}$

The moles of $\left[C{u}^{2+}\right]$left in the given solution $=4.00-3.00$

Concentration of $\left[C{u}^{2+}\right]$ions$=\frac{\text{1.00}Mol}{4.00Litre}$

$=0.250mol$

$\left[N{O}_3^{-}\right]=\frac{\text{8.00}Mol}{4.00Litre}$

$=2.00\raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right.$

$\left[K^{+}\right]=\frac{\text{6.00}Mol}{4.00Litre}$

$=1.5\raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right.$

Therefore the left over concentrations of the solutionare.$\left[C{u}^{2+}\right]=0.250mol$