Q13.59 P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molarity of each aqueous solution: 

(a) 25.5 mL of 6.25 M HCl diluted to 0.500 L with water 

(b) 8.25 mL of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}$ KI diluted to 12.0 mL with water

Step-by-Step Solution

Answer

1. The molarity of the HCl is 0.32M after dilution.
2. The molarity of the KI is 0.1375M after dilution.

Step 1: Concentration of the Solution

Solution can be formed from the dissolution of the solute in the solvent. 

The number of moles of solute remains unchanged after dilution. The number of moles of solute is the product of molarity of the solution and the volume of the solution.

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

Number of moles can be defined as the ratio of the mass of the atom/molecule and molar mass of the atom/molecule.

$\mathbf{Number}\mathbf{ } \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ } \mathbf{Mass}}$

Step 2: Subpart (a)

The final volume of  HCl solution is  $V_2=0.500 L$

Initial molarity of HCl at  $M_1=6.25 M$

Initial volume is  $V_1=0.025 mL$

Molarity of solution after dilution be  $M_2$.

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

$$\begin{gathered} \mathbf{6}\mathbf{.}\mathbf{25}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0255}\mathbf{L}\mathbf{=}{\mathbf{M}}_{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{L} \\ {\mathbf{M}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{25}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0255}\mathbf{L}}{\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{L}} \\ {\mathbf{M}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{32}\mathbf{M} \end{gathered}$$

Step 3: Subpart (b)

Initial volume of the solution is    $V_1=8.25 mL=0.00825 L$

Initial molarity of KI solution 

$$\begin{gathered} M_1=2.00\times {10}^{-2} M \\ =0.02 M \end{gathered}$$

Volume of Solution after dilution is  $V_2=0.012 L$

Molarity of solution after dilution be  $M_2$.

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

$$\begin{gathered} \mathbf{0}\mathbf{.}\mathbf{02}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0}0\mathbf{825}\mathbf{L}\mathbf{=}{\mathbf{M}}_{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{012}\mathbf{L} \\ {\mathbf{M}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0}0\mathbf{825}\mathbf{L}}{\mathbf{0}\mathbf{.}\mathbf{012}\mathbf{L}} \\ {\mathbf{M}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1375}\mathbf{M} \end{gathered}$$