Q. 9.59 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine the initial volume, in milliliter, requires preparing each of the following:

a. $255 \mathrm{mL} \mathrm{of} \mathrm{a} 0.200 \mathrm{M} {\mathrm{HNO}}_3$ solution using a$4.00 \mathrm{M} {\mathrm{HNO}}_3$ solution

b. $715 \mathrm{mL} \mathrm{of} \mathrm{a} 0.100 \mathrm{M} {\mathrm{MgCl}}_2$solution using a $6.00 \mathrm{M} {\mathrm{MgCl}}_2$ solution

c. $0.100 \mathrm{L} \mathrm{of} \mathrm{a} 0.150 \mathrm{M} \mathrm{KCl}$ solution using an $8.00 \mathrm{M} \mathrm{KCl}$solution

Step-by-Step Solution

Answer

a. The initial volume of $255 \mathrm{mL} \mathrm{of} \mathrm{a} 0.200 \mathrm{M} {\mathrm{HNO}}_3$ solution using an $4.00 \mathrm{M} {\mathrm{HNO}}_3$is $12.75 \mathrm{mL}$.

b. The initial volume of $715 \mathrm{mL} \mathrm{of} \mathrm{a} 0.100 \mathrm{M} {\mathrm{MgCl}}_2$ solution using an $6.00 \mathrm{M} {\mathrm{MgCl}}_2$ is $11.9 \mathrm{mL}$.

c. The initial volume of $0.100 \mathrm{L} \mathrm{of} \mathrm{a} 0.150 \mathrm{M} \mathrm{KCl}$ solution using an $8.00 \mathrm{M} \mathrm{KCl}$  is $1.875 \mathrm{mL}$.

Part (a) Step 1: Given Information

We need to find the initial volume.

Part (a) Step 2: Explanation

The addition of solvent increases the volume, and the amount of solute does not change. The concentrated solution or the diluted solution. Moles or grams  of solute$=$moles or grams   of solute

$={\mathrm{C}}_1{\mathrm{V}}_1={\mathrm{C}}_2{\mathrm{V}}_2$

Consider:

$$\begin{gathered} {\mathrm{C}}_1=4 \mathrm{M} \\ {\mathrm{C}}_2=0.2 \mathrm{M} \\ {\mathrm{V}}_2=255 \mathrm{mL} \end{gathered}$$

Now, finding the value of $V_1$:

$$\begin{gathered} {\mathrm{V}}_1=\frac{{\mathrm{C}}_2{\mathrm{V}}_2 }{{\mathrm{C}}_1} \\ =\frac{(0.2 \mathrm{M})(255 \mathrm{mL})}{4 \mathrm{M}} \\ =12.75 \mathrm{mL} \end{gathered}$$

Part (b) Step 1: Given Information

We need to find the initial volume.

Part (b) Step 2: Explanation

The addition of solvent increases the volume, and the amount of solute does not change. The concentrated solution or the diluted solution. Moles or grams  of solute$=$moles or grams  of solute

$={\mathrm{C}}_1{\mathrm{V}}_1={\mathrm{C}}_2{\mathrm{V}}_2$

Consider:

$$\begin{gathered} {\mathrm{C}}_1=6 \mathrm{M} \\ {\mathrm{C}}_2=0.1 \mathrm{M} \\ {\mathrm{V}}_2=715 \mathrm{mL} \end{gathered}$$

Now, finding the value of $V_1$:

$$\begin{gathered} {\mathrm{V}}_1=\frac{{\mathrm{C}}_2{\mathrm{V}}_2 }{{\mathrm{C}}_1} \\ =\frac{(0.1\mathrm{M})(715 \mathrm{mL})}{6 \mathrm{M}} \\ =11.09 \mathrm{mL} \end{gathered}$$

Part (c) Step 1: Given Information

We need to find the initial volume.

Part (c) Step 2: Explanation

Here,

$={\mathrm{C}}_1{\mathrm{V}}_1={\mathrm{C}}_2{\mathrm{V}}_2$

Consider:

$$\begin{gathered} {\mathrm{C}}_1=0.150 \mathrm{M} \\ {\mathrm{C}}_2=8 \mathrm{M} \\ {\mathrm{V}}_2=0.100 \mathrm{L}=100 \mathrm{mL} \end{gathered}$$

Now, finding the value of $V_1$:

$$\begin{gathered} {\mathrm{V}}_1=\frac{{\mathrm{C}}_2{\mathrm{V}}_2 }{{\mathrm{C}}_1} \\ =\frac{(0.150 \mathrm{M})(100 \mathrm{mL})}{8 \mathrm{M}} \\ =1.875 \mathrm{mL} \end{gathered}$$