Chapter 15, Problem 38CR

Interpretation Introduction

(a)

Interpretation:

The mass of pure ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ present in pure $125\text{mL}$ sample is to be calculated.

Concept Introduction:

The mass percentage of a compound present in a solution is calculated by dividing the mass of the compound present in the solution by the total mass of the solution and then multiplying it by $100$. The formula for mass percentage is represented as,

$\text{mass}\text{\%}=\frac{\text{M}}{{\text{M}}_{\text{t}}}\times 100\text{\%}$

Where,

• $\text{M}$ represents the mass of the compound present in the solution.
• ${\text{M}}_{\text{t}}$ represents the total mass of solution.

Answer to Problem 38CR

The mass of pure ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ present in $125\text{mL}$ sample of concentrated sulfuric acid solution is $226.09\text{g}$.

Explanation of Solution

The volume of the concentrated sulfuric acid solution is $125\text{mL}$.

The density of the concentrated sulfuric acid solution is $1.84\text{g}/\text{mL}$.

The mass percentage of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given solution is $98.3\text{\%}$.

The relation between mass, volume and density of a substance is given as,

$\text{m}=\text{V}\text{D}$

Where,

• $\text{V}$ represents the volume occupied by the substance.
• $\text{m}$ represents the mass of the substance.
• $\text{D}$ represents the density of the substance.

Substitute the value of density and volume of sulfuric acid solution in the above equation.

$\begin{array}{c}\text{m}=\left(125\text{mL}\right)\left(1.84\text{g}/\text{mL}\right)\\ =230.00\text{g}\end{array}$

The formula for mass percentage is represented as,

$\text{mass}\text{\%}=\frac{\text{M}}{{\text{M}}_{\text{t}}}\times 100\text{\%}$

Where,

• $\text{M}$ represents the mass of the compound present in the solution.
• ${\text{M}}_{\text{t}}$ represents the total mass of solution.

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

$\text{M}=\frac{{\text{M}}_{\text{t}}\times \text{mass}\text{\%}}{100\text{\%}}$

Substitute the value of mass of concentrated sulfuric acid solution and mass percentage of sulfuric acid in the above equation.

$\begin{array}{c}\text{M}=\frac{\left(230.00\text{g}\right)\left(98.3\text{\%}\right)}{100\text{\%}}\\ =226.09\text{g}\end{array}$

Therefore, the mass of pure ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ present in $125\text{mL}$ sample of concentrated sulfuric acid solution is $226.09\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The molarity of given concentrated acid solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of mole 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 38CR

The molarity of the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $18.44\text{M}$.

Explanation of Solution

The mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ taken is $226.09\text{g}$.

The volume of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $125\text{mL}$.

The molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $98.079\text{g}\cdot {\text{mol}}^{-1}$.

The molarity of a solution is given as,

$\text{M}=\frac{\text{m}}{{\text{M}}_{\text{m}}\text{V}}$

Where,

• $\text{m}$ represents the mass of the solute.
• $\text{V}$ represents the volume of the solution.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the solute.

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

$\begin{array}{c}\text{M}=\frac{\left(226.09\text{g}\right)\left(\frac{1\text{M}}{1\text{mol}\cdot {\text{L}}^{-1}}\right)}{\left(98.07\text{g}\cdot {\text{mol}}^{-1}\right)\left(125\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)}\\ =18.44\text{M}\end{array}$

Therefore, the molarity of the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $18.44\text{M}$.

Interpretation Introduction

(c)

Interpretation:

The molarity of the given acid solution after dilution 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 38CR

The molarity of the given acid solution after dilution is $0.766\text{M}$.

Explanation of Solution

The initial molarity of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $18.44\text{M}$.

The initial volume of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $125\text{mL}$.

The final volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $3.01\text{L}$.

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 ${\text{M}}_{\text{f}}$.

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

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

$\begin{array}{c}{\text{M}}_{\text{f}}=\frac{\left(18.44\text{M}\right)\left(125\text{mL}\right)}{\left(3.01\text{L}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)}\\ =0.766\text{M}\end{array}$

Therefore, the molarity of the given acid solution after dilution is $0.766\text{M}$.

Interpretation Introduction

(d)

Interpretation:

The normality of the dilute acid solution is to be calculated.

Concept Introduction:

The normality of a solution is defined as the gram equivalent weight of a solute dissolved in one liter of the solution. The normality of solution is given as:

$\text{N}=\frac{{\text{n}}_{\text{eq}}}{\text{V}}$

Where,

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

Answer to Problem 38CR

The normality of the diluted ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $1.532\text{N}$.

Explanation of Solution

The molarity of the diluted solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $0.766\text{M}$.

The volume of $0.766\text{M}{\text{H}}_2{\text{SO}}_{\text{4}}$ solution is $3.01\text{L}$.

The dissociation of sulfuric acid in water is represented as:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{a}\text{q}\right)\to 2{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)$

Therefore, the equivalence factor of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $2\text{eq}\cdot {\text{mol}}^{-1}$.

The relationship between the molarity and normality of a solution is given as,

$\text{N}=\text{n}\text{'}\text{M}$

Where,

• $\text{N}$ represents the normality of the solution.
• $\text{M}$ represents the molarity of the solution.
• $\text{n}\text{'}$ represents the equivalence factor of the solute.

Substitute the value of $\text{M}$ and $\text{n}\text{'}$ in the above equation.

$\begin{array}{c}\text{N}=\left(2\text{eq}\cdot {\text{mol}}^{-1}\right)\left(0.766\text{M}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\left(\frac{1\text{N}}{1\text{eq}\cdot {\text{L}}^{-1}}\right)\\ =1.532\text{N}\end{array}$

Therefore, the normality of the diluted ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $1.532\text{N}$.

Interpretation Introduction

(e)

Interpretation:

The volume of dilute acid solution required to neutralize $45.3\text{mL}$ of $0.532\text{M}\text{NaOH}$ solution is to be calculated.

Concept Introduction:

Neutralization is the process in which an acid and base react with each other to give salt and water. The general reaction of acid and base is represented as,

$\text{AH}+\text{BOH}\to \text{AB}+{\text{H}}_{\text{2}}\text{O}$

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 38CR

The volume of dilute acid solution required to neutralize $36.2\text{mL}$ of $0.532\text{M}\text{NaOH}$ is $15.7308\text{mL}$.

Explanation of Solution

The molarity of the dilute ${\text{H}}_2{\text{SO}}_4$ solution is $0.766\text{M}$.

The molarity of the given $\text{NaOH}$ solution is $0.532\text{M}$.

The volume of the $0.532\text{M}\text{NaOH}$ is $45.3\text{mL}$.

The neutralization reaction between ${\text{H}}_2{\text{SO}}_4$ and $\text{NaOH}$ is represented as,

${\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)+2\text{NaOH}\left(\text{a}\text{q}\right)\to {\text{Na}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The relation between molarities of ${\text{H}}_2{\text{SO}}_4$ and $\text{NaOH}$ is given as,

$2{\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$ represents the molarity of ${\text{H}}_2{\text{SO}}_4$.
• ${\text{V}}_1$ represents the volume of ${\text{H}}_2{\text{SO}}_4$.
• ${\text{M}}_2$ represents the molarity of $\text{NaOH}$.
• ${\text{V}}_2$ represents the volume of $\text{NaOH}$.

Rearrange the above equation for the value of ${\text{V}}_1$.

${\text{V}}_1=\frac{{\text{M}}_2{\text{V}}_2}{2{\text{M}}_1}$

Substitute the value ${\text{M}}_1$, ${\text{M}}_2$ and ${\text{V}}_2$ in the above equation.

$\begin{array}{c}{\text{V}}_1=\frac{\left(0.532\text{M}\right)\left(45.3\text{mL}\right)}{\left(2\right)\left(0.766\text{M}\right)}\\ =15.7308\text{mL}\end{array}$

Therefore, the volume of dilute acid solution required to neutralize $36.2\text{mL}$ of $0.532\text{M}\text{NaOH}$ is $15.7308\text{mL}$.