Q. 9.60 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a.  $20.0 \mathrm{mL} \mathrm{of} \mathrm{a} 0.250 \mathrm{M} {\mathrm{KNO}}_3$ solution using a $6.00 \mathrm{M} {\mathrm{KNO}}_3$ solution

b. $25.0 \mathrm{mL} \mathrm{of} \mathrm{a} 2.50 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ solution using a $12.0 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$solution

c. $0.500 \mathrm{L} \mathrm{of} \mathrm{a} 1.50 \mathrm{M} {\mathrm{NH}}_4\mathrm{Cl}$ solution using a $10.0 \mathrm{M} {\mathrm{NH}}_4\mathrm{Cl}$ solution

Step-by-Step Solution

Answer

a. The initial volume of $20.0 \mathrm{mL} \mathrm{of} \mathrm{a} 0.250 \mathrm{M} {\mathrm{KNO}}_3$ solution using an $6.00 \mathrm{M} {\mathrm{KNO}}_3$ is $0.83 \mathrm{mL}$.

b. The initial volume of $25.0 \mathrm{mL} \mathrm{of} \mathrm{a} 2.50 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ solution using an$12.0 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ is $5.21 \mathrm{mL}$.

c. The initial volume of  $0.500 \mathrm{L} \mathrm{of} \mathrm{a} 1.50 \mathrm{M} {\mathrm{NH}}_4\mathrm{Cl}$ solution using an $10.0 \mathrm{M} {\mathrm{NH}}_4\mathrm{Cl}$ is$75 \mathrm{mL}$.

Part (a) Step 1: Given Information

$20.0\mathrm{mL}\text{ of a }0.250\mathrm{M}{\mathrm{KNO}}_3\text{ solution using a }6.00\mathrm{M}{\mathrm{KNO}}_3\text{ solution }$

Part (a) Step 2: Explanation

We know,

$C_1{V}_1=C_2{V}_2$

Now 

${\mathrm{C}}_1=6\mathrm{M} {\mathrm{C}}_2=0.25\mathrm{M} {\mathrm{V}}_2=20\mathrm{mL}$

The initial volume is

$V_1=\frac{C_2{V}_2}{C_1}=\frac{0.25\mathrm{M}\times 20\mathrm{mL}}{6\mathrm{M}}=0.83\mathrm{mL}$

Part (b) step 1: Given Information

$25.0\mathrm{mL}\text{ of a }2.50{\mathrm{MH}}_2{\mathrm{SO}}_4\text{ solution using a }12.0{\mathrm{MH}}_2{\mathrm{SO}}_4\text{ solution }$

Part (b) Step 2: Explanation

We know,

$C_1{V}_1=C_2{V}_2$

Now,

$\begin{array}{r}{C}_1=12\mathrm{M} {\mathrm{C}}_2=2.5\mathrm{M} {\mathrm{V}}_2=25\mathrm{mL}\end{array}$

The initial volume is

$\begin{array}{r}{V}_1=\frac{C_2{V}_2}{C_1}=\frac{2.5\mathrm{M}\times 25\mathrm{mL}}{12\mathrm{M}}=5.21\mathrm{mL}\end{array}$

Part (c) Step 1: Given Information

$0.500\mathrm{L}\text{ of a }1.50\mathrm{M}{\mathrm{NH}}_4\mathrm{Cl}\text{ solution using a }10.0\mathrm{M}{\mathrm{NH}}_4\mathrm{Cl}\text{ solution }$

Part (c) Step 2: Explanation

We know,

$C_1{V}_1=C_2{V}_2$

Now,

$\begin{array}{r}{C}_1=10\mathrm{M} {\mathrm{C}}_2=1.5\mathrm{M} {\mathrm{V}}_2=0.5\mathrm{L}=500\mathrm{mL}\end{array}$

The initial volume is

$\begin{array}{r}{V}_1=\frac{C_2{V}_2}{C_1}=\frac{1.5\mathrm{M}\times 500\mathrm{mL}}{10\mathrm{M}}=75\mathrm{mL}\end{array}$