Chapter 15, Problem 122AP

Interpretation Introduction

(a)

Interpretation:

The new molarity of the given solution diluted with $150\text{mL}$ of water is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 122AP

The new molarity of the $0.200\text{M}\text{HBr}$ solution on dilution is $0.091\text{M}$.

Explanation of Solution

The molarity of the given $\text{HBr}$ solution is $0.200\text{M}$.

The volume of $0.200\text{M}\text{HBr}$ solution is $125\text{mL}$.

The volume of water added in the solution is $150\text{mL}$.

The total new volume of the solution is given as:

${\text{V}}_{\text{f}}={\text{V}}_{\text{i}}+{\text{V}}_{\text{w}}$

Where,

• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.
• ${\text{V}}_{\text{w}}$ represents the volume of water added in the solution.

Substitute the value of ${\text{V}}_{\text{i}}$ and ${\text{V}}_{\text{w}}$ in the above equation.

$\begin{array}{c}{\text{V}}_{\text{f}}=125\text{mL}+150\text{mL}\\ =275\text{mL}\end{array}$

The relation between the initial and final volume of a solution is given as:

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Where,

• ${\text{M}}_{\text{f}}$ represents the final molarity of the solution.
• ${\text{V}}_{\text{f}}$ represents the final volume of the solution.
• ${\text{M}}_{\text{i}}$ represents the initial molarity of the solution.
• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.

Rearrange the above equation for the value of ${\text{M}}_{\text{f}}$.

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Substitute the value of ${\text{V}}_{\text{f}}$, ${\text{M}}_{\text{i}}$ and ${\text{V}}_{\text{i}}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{f}}=\frac{\left(0.200\text{M}\right)\left(125\text{mL}\right)}{275\text{mL}}\\ =0.091\text{M}\end{array}$

Therefore, the new molarity of the $0.200\text{M}\text{HBr}$ solution on dilution is $0.091\text{M}$.

Interpretation Introduction

(b)

Interpretation:

The new molarity of the given solution diluted with $150\text{mL}$ of water is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 122AP

The new molarity of the $0.250\text{M}\text{Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2$ solution on dilution is $0.127\text{M}$.

Explanation of Solution

The molarity of the given $\text{Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2$ solution is $0.250\text{M}$.

The volume of $0.250\text{M}\text{Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2$ solution is $155\text{mL}$.

The volume of water added in the solution is $150\text{mL}$.

The total new volume of the solution is given as:

${\text{V}}_{\text{f}}={\text{V}}_{\text{i}}+{\text{V}}_{\text{w}}$

Where,

• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.
• ${\text{V}}_{\text{w}}$ represents the volume of water added in the solution.

Substitute the value of ${\text{V}}_{\text{i}}$ and ${\text{V}}_{\text{w}}$ in the above equation.

$\begin{array}{c}{\text{V}}_{\text{f}}=155\text{mL}+150\text{mL}\\ =305\text{mL}\end{array}$

The relation between the initial and final volume of a solution is given as:

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Where,

• ${\text{M}}_{\text{f}}$ represents the final molarity of the solution.
• ${\text{V}}_{\text{f}}$ represents the final volume of the solution.
• ${\text{M}}_{\text{i}}$ represents the initial molarity of the solution.
• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.

Rearrange the above equation for the value of ${\text{M}}_{\text{f}}$.

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Substitute the value of ${\text{V}}_{\text{f}}$, ${\text{M}}_{\text{i}}$ and ${\text{V}}_{\text{i}}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{f}}=\frac{\left(0.250\text{M}\right)\left(155\text{mL}\right)}{305\text{mL}}\\ =0.127\text{M}\end{array}$

Therefore, the new molarity of the $0.250\text{M}\text{Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2$ solution on dilution is $0.127\text{M}$.

Interpretation Introduction

(c)

Interpretation:

The new molarity of the given solution diluted with $150\text{mL}$ of water is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 122AP

The new molarity of the $0.250\text{M}{\text{H}}_3{\text{PO}}_4$ solution on dilution is $0.192\text{M}$.

Explanation of Solution

The molarity of the given ${\text{H}}_3{\text{PO}}_4$ solution is $0.250\text{M}$.

The volume of $0.250\text{M}{\text{H}}_3{\text{PO}}_4$ solution is $0.500\text{L}$.

The volume of water added in the solution is $150\text{mL}$.

The total new volume of the solution is given as:

${\text{V}}_{\text{f}}={\text{V}}_{\text{i}}+{\text{V}}_{\text{w}}$

Where,

• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.
• ${\text{V}}_{\text{w}}$ represents the volume of water added in the solution.

Substitute the value of ${\text{V}}_{\text{i}}$ and ${\text{V}}_{\text{w}}$ in the above equation.

$\begin{array}{c}{\text{V}}_{\text{f}}=\left(0.500\text{L}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)+150\text{mL}\\ =650\text{mL}\end{array}$

The relation between the initial and final volume of a solution is given as:

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Where,

• ${\text{M}}_{\text{f}}$ represents the final molarity of the solution.
• ${\text{V}}_{\text{f}}$ represents the final volume of the solution.
• ${\text{M}}_{\text{i}}$ represents the initial molarity of the solution.
• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.

Rearrange the above equation for the value of ${\text{M}}_{\text{f}}$.

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Substitute the value of ${\text{V}}_{\text{f}}$, ${\text{M}}_{\text{i}}$ and ${\text{V}}_{\text{i}}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{f}}=\frac{\left(0.250\text{M}\right)\left(0.500\text{L}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)}{650\text{mL}}\\ =0.192\text{M}\end{array}$

Therefore, the new molarity of the $0.250\text{M}{\text{H}}_3{\text{PO}}_4$ solution on dilution is $0.192\text{M}$.

Interpretation Introduction

(d)

Interpretation:

The new molarity of the given solution diluted with $150\text{mL}$ of water is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 122AP

The new molarity of the $18.0\text{M}{\text{H}}_2{\text{SO}}_4$ solution on dilution is $1.636\text{M}$.

Explanation of Solution

The molarity of the given ${\text{H}}_2{\text{SO}}_4$ solution is $18.0\text{M}$.

The volume of $18.0\text{M}{\text{H}}_2{\text{SO}}_4$ solution is $15\text{mL}$.

The volume of water added in the solution is $150\text{mL}$.

The total new volume of the solution is given as:

${\text{V}}_{\text{f}}={\text{V}}_{\text{i}}+{\text{V}}_{\text{w}}$

Where,

• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.
• ${\text{V}}_{\text{w}}$ represents the volume of water added in the solution.

Substitute the value of ${\text{V}}_{\text{i}}$ and ${\text{V}}_{\text{w}}$ in the above equation.

$\begin{array}{c}{\text{V}}_{\text{f}}=15\text{mL}+150\text{mL}\\ =165\text{mL}\end{array}$

The relation between the initial and final volume of a solution is given as:

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Where,

• ${\text{M}}_{\text{f}}$ represents the final molarity of the solution.
• ${\text{V}}_{\text{f}}$ represents the final volume of the solution.
• ${\text{M}}_{\text{i}}$ represents the initial molarity of the solution.
• ${\text{V}}_{\text{i}}$ represents the initial volume of the solution.

Rearrange the above equation for the value of ${\text{M}}_{\text{f}}$.

${\text{M}}_{\text{f}}{\text{V}}_{\text{f}}={\text{M}}_{\text{i}}{\text{V}}_{\text{i}}$

Substitute the value of ${\text{V}}_{\text{f}}$, ${\text{M}}_{\text{i}}$ and ${\text{V}}_{\text{i}}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{f}}=\frac{\left(18.0\text{M}\right)\left(15\text{mL}\right)}{165\text{mL}}\\ =1.636\text{M}\end{array}$

Therefore, the new molarity of the $18.0\text{M}{\text{H}}_2{\text{SO}}_4$ solution on dilution is $1.636\text{M}$.