Q13.60 P - Chemistry: Molecular Nature Of Matter And Change

Question

How would you prepare the following aqueous solutions? 

(a) 365 mL of  $\mathbf{8}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}$  ${\mathbf{KH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$ from solid  ${\mathbf{KH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$

(b) 465 mL of 0.335 M NaOH from 1.25 M NaOH.

Step-by-Step Solution

Answer

a) 4.22 g of the solid $K{H}_{2}P{O}_4$ should be dissolved in 365 mL water.

 

b) Take 125 mL of 1.25 M NaOH solution and add water until the volume becomes 465 mL.

Step 1: The mass of solute required from molarity

By using the molarity formula, the number of moles of solute required can be calculated.

$\mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Volume} \mathbf{of}\mathbf{ } \mathbf{the}\mathbf{ } \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}}$

Then, multiply the number of moles of solute with its molar mass. 

 

When a solution is diluted, the number of moles of the solute remains the same.

The number of moles of solute can be obtained by:

$n=M\times V$

 

Where, M is the molarity of solution and V is the volume of solution.

To find the molarity of a solution after dilution, use the formula:

$M_1{V}_1=M_2{V}_2$

  

Step 2: Subpart (a)

Molarity of the  $\mathbf{KH}\mathbf{2}{\mathbf{PO}}_{\mathbf{4}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}$

 $$\begin{gathered} \mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Solution}\mathbf{=}\mathbf{365}\mathbf{mL} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{365}\mathbf{L} \end{gathered}$$

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{0}\mathbf{.}\mathbf{0855} \mathbf{M}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{0}\mathbf{.}\mathbf{365} \mathbf{L}} \\ \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0855}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{365} \mathbf{mol} \\ \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{031}\mathbf{moles} \end{gathered}$$

The mass of  $K{H}_{2}P{O}_4$ solid is:

$$\begin{gathered} mass=n\times M \\ =0.031 mol\times 136.1 g/mol \\ =4.22 g \end{gathered}$$

Hence, 4.22 g of the solid $K{H}_{2}P{O}_4$ should be dissolved in 365 mL water.

Step 3: Subpart (b)

Initial molarity  $M_1=1.25 M$

Final molarity  $M_2=0.335 M$

Final volume  $V_2=465 mL$

$$\begin{gathered} \mathbf{Molarity} \mathbf{Equation}\mathbf{,} \\ {\mathbf{M}}_{\mathbf{1}}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}}\mathbf{=}{\mathbf{M}}_{\mathbf{2}}\mathbf{\times }{\mathbf{V}}_{\mathbf{2}} \end{gathered}$$

$$\begin{gathered} \mathbf{0}\mathbf{.}\mathbf{335}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{465}\mathbf{L}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{M}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}} \\ {\mathbf{V}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{335}\mathbf{M}\mathbf{\times }\mathbf{465} \mathbf{mL}}{\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{M}} \\ {\mathbf{V}}_{\mathbf{1}}\mathbf{=}\mathbf{125} \mathbf{mL} \end{gathered}$$

Take 125 mL of 1.25 M NaOH solution and add water until the volume becomes 465 mL.