Question

What volume of 0.100 MNaOH is required to precipitate all of the nickel(II) ions from 150.0 mL of a 0.249-Msolution of Ni(NO₃)₂?

Short answer

Answer:

The volume of solution that evaporated is 747 mL.

Step-by-step solution

Step by Step Solution Step 1: Definition of molarity of a solution.

Molarity which is basically the concentration can be calculated by fractionating the number of moles with the volume of solution in liter.

It can be formulated as:

$\text{Molarity}\left(M{\text{or molL}}^{-1}\right)=\frac{\text{Number of moles}\left(n\right)}{\text{Volume of solution}\left(\text{in L}\right)}\cdots \cdots \left(1\right)$

The balanced chemical reaction of nickel (II) ions of Ni(NO3)2.

$\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}+\text{2NaOH}\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}+{\text{2NaNO}}_{\text{3}}$

Conversion of volume from mL to L.

In formula of molarity, the volume of solution is present in L, thus 150 mL can be converted to L as:

$$\begin{gathered} \text{1 L = 1000 mL} \\ \text{Conversion factor =}\frac{\text{1 L}}{\text{1000 mL}} \\ \text{Volume}\left(\text{in L}\right)\text{= Volume}\left(\text{in mL}\right)\times \text{Conversion factor} \\ \text{Volume}\left(\text{in L}\right)\text{= 150}\times \frac{\text{1 L}}{\text{1000}} \\ =\text{0.150 L}\underset{}{\to } \end{gathered}$$

Number of moles present in 0.150 L and 0.249 M of nickel solution.

Number of moles can be calculated by rearranging equation (1) as:

$$\begin{gathered} \text{Number of moles}\left(n\right)=\text{Molarity}\times \text{Volume} \\ \text{= 0.249 mol}\times 0.150 \\ =\text{0.0373 mol} \end{gathered}$$

Sincetwo moles of hydroxide ion stochiometryare used in the chemical reaction, the number of moles of hydroxide ion will be:

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\left(n\right)=2\times 0.0373\mathrm{mol} \\ =0.0747\mathrm{mol} \end{gathered}$$

Calculating the final volume of solution.

Since the number of moles of the solution is same before and after heating on hot plate, the volume of the solution can be calculated by rearranging equation (1) as:

$$\begin{gathered} \text{Volume of solution}\left(\text{in L}\right)\text{=}\frac{\text{Number of moles}\left(n\right)}{\text{Molarity}\left({\text{molL}}^{-\text{1}}\right)} \\ \text{=}\frac{\text{0.0747}}{{\text{0.100L}}^{-\text{1}}} \\ =\text{0.747 L} \end{gathered}$$

Conversion of volume from L to mL.

$$\begin{gathered} \text{1 L = 1000 mL} \\ \text{Conversion factor =}\frac{\text{1000 mL}}{\text{1 L}} \\ \text{Volume}\left(\text{in mL}\right)\text{= Volume}\left(\text{in L}\right)\times \text{Conversion factor} \\ \text{Volume}\left(\text{in mL}\right)\text{= 0.747}\times \frac{\text{1000 mL}}{\text{1}} \\ =747\text{mL} \end{gathered}$$

Thus, the volume of solution that evaporated is 747mL.