Chapter 15, Problem 41A

(a)

Interpretation Introduction

Interpretation:

The volume of the $1.0\text{0M}\text{NaOH}$ solution required to neutralize the given acetic acid solution is to be calculated.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V\left(\text{L}\right)}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions

Answer to Problem 41A

  $\text{3}\text{.85mL}$

Explanation of Solution

Given information:

The acetic acid volume is, $25.\text{0mL}$

The molarities of acetic acid and sodium hydroxide is,

$M_1=1.\text{00M}$ and $M_2=0.15\text{4M}$

Calculation:

The molarity formula is,

  $M\text{=}\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}$

And,

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

The balanced equation of the given solution is,

  ${\mathrm{CH}}_{3}COOH\left(aq\right)+\text{NaOH}\left(aq\right)\to {\text{CH}}_{\text{3}}\text{COONa}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}$

Hence, 1 mole of ${\text{CH}}_{\text{3}}\text{COOH}$ is required to neutralize 1 mole sodium hydroxide solution.

So, Firstly, calculate the number of moles of hydrogen ion in $\text{25}\text{.0mL}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}_{\text{2}}$ solution such that,

  $\begin{array}{l}n=M\times V\\ =\left(\text{25}{\text{.0×10}}^{\text{-3}}\text{L}\right)\text{×}\left(\text{0}\text{.154M}\right)\\ \text{=}\text{3}{\text{.85×10}}^{\text{-3}}\text{mol}\end{array}$

For neutralizing the reaction, the number of hydrogen ions must be equal to the number of hydroxide ions. The hydroxide ions are limiting.

According o the balanced equations both hydrogen and hydroxide ions are reacted in a $1:1$ ratio.

Therefore, the number of moles of hydroxide ions are,

  $n_{{\text{OH}}^{\text{-}}}=3.85\times {10}^{-3}\text{mol}$

Now, calculate the volume of the $\text{1}\text{.00M}\text{NaOH}$ solution is,

  $\begin{array}{l}n=M\times V\\ \Rightarrow V_{O{H}^{-}}=\frac{n}{M}\\ =\frac{\text{3}{\text{.85×10}}^{\text{-3}}\text{mol}}{\text{1}\text{.00}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}\\ \text{ =}\text{3}{\text{.85×10}}^{\text{-3}}\text{mol}\end{array}$

The volume of the $\text{1}\text{.00M}\text{NaOH}$ solution to neutralized the $\text{25}\text{.0mL}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}_{\text{2}}$ solution is $\text{3}\text{.85mL}$ .

(b)

Interpretation Introduction

Interpretation:

The volume of the $\text{1}\text{.00M}\text{NaOH}$ solution required to neutralize the given HF solution is to be calculated.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V\left(\text{L}\right)}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions

Answer to Problem 41A

  $3.5\text{7mL}$

Explanation of Solution

Given information:

The volume is, $V=35.\text{0mL}$

The molarities of $\mathrm{HF}$ and sodium hydroxide is,

$M_1=1.\text{00M}$ and $M_2=0.102M$

Calculation:

• The molarity formula is,

  $M=\mathrm{molarity}=\frac{numberofmolesofsolute}{volumeofthesolution}$

And,

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

The balanced equation of the given solution is,

  $\text{HF}\left(\text{aq}\right)\text{+NaOH}\left(\text{aq}\right)\to \text{NaF}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}$

Hence, 1 mole of $HF$ is required to neutralize 1 mole $NaOH$ solution.

So, Firstly, calculate the number of moles of hydrogen ion in solution $HF$ such that,

  $\begin{array}{l}{n}_{H^{+}}=M\times V\text{=}\left(\text{0}\text{.102}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\right)\text{×}\left(\text{35}{\text{.0×10}}^{\text{-3}}\text{L}\right)\\ \text{=}\text{3}{\text{.57×10}}^{\text{-3}}\text{mol}\end{array}$

For neutralizing the reaction, the number of hydrogen ions must be equal to the number of hydroxide ions. The hydroxide ions are limiting.

According to the balanced equations both hydrogen and hydroxide ions are reacted in a $1:1$ ratio.

Therefore, the number of moles of hydroxide ions are,

  $n_{O{H}^{-}}=3.57\times {10}^{-3}\text{mol}$

Now, calculate the volume of the $1.00MNaOH$ solution is,

$\begin{array}{l}{n}_{O{H}^{-}}=\text{3}{\text{.57×10}}^{\text{-3}}\text{mol}\\ n=M\times V\\ V_{O{H}^{-}}=\frac{n}{M}\\ =\frac{\text{3}{\text{.57×10}}^{\text{-3}}\text{mol}}{\text{1}\text{.00}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}\\ \text{=}\text{3}{\text{.57×10}}^{\text{-3}}\text{mol}\end{array}$

The volume of the $\text{1}\text{.00M}\text{NaOH}$ solution to neutralized the $35.0\text{mL}\text{HF}$ solution is $\text{3}\text{.57mL}$ .

(c)

Interpretation Introduction

Interpretation:

The volume of the $\text{1}\text{.00M}\text{NaOH}$ solution required to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is to be calculated.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V\left(\text{L}\right)}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions

Answer to Problem 41A

  $\text{4}\text{.29mL}$

Explanation of Solution

Given information:

The volume is, $V=10.\text{0mL}$

The molarities of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and sodium hydroxide is,

$M_1=1.00M$ and $M_2=\text{0}\text{.143M}$

Calculation:

• The molarity formula is,

  $M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}$

And,

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

The balanced equation of the given solution is,

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{aq}\right)\text{+3NaOH}\left(\text{aq}\right)\to {\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{aq}\right){\text{+3H}}_{\text{2}}\text{O}$

Hence, 1 mole of $H_{3}P{O}_4$ is required to neutralize 3 moles $NaOH$ solution Firstly, calculate the number of moles of hydrogen ion in $10.0mL{H}_{3}P{O}_4$ solution such that,

For neutralizing the reaction, the number of hydrogen ions must be equal to the number of hydroxide ions.

According to the balanced equations both hydrogen and hydroxide ions are reacted in a $1:3$ ratio.

Therefore, the number of moles of hydroxide ions are,

  $\text{0}\text{.00143moles×}\frac{\text{3}\text{moles}\text{NaOH}}{\text{1mole}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=0}\text{.00429moles}$

Now, calculate the volume of the $\text{1}\text{.00M}\text{NaOH}$ solution for 1 mole hydroxide ion is,

  $\begin{array}{l}n=M\times V\\ \Rightarrow V_{O{H}^{-}}=\frac{n}{M}\\ =\frac{\text{0}\text{.00429}\text{moles}}{\text{1}\text{.00}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}\text{=}\text{4}\text{.29mL}\end{array}$

The volume of $\text{1}\text{.00M}\text{NaOH}$ solution is $\text{4}\text{.29mL}$ .

(d)

Interpretation Introduction

Interpretation:

The volume of the $1.00MNaOH$ solution, which is required to neutralize the given solution, is to be calculated.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V\left(\text{L}\right)}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions

.

Answer to Problem 41A

  $\text{15}\text{.4mL}$

Explanation of Solution

Given information:

The volume is, $V=35.\text{0mL}$

The molarities of $0.220{\text{MH}}_{\text{2}}{\text{SO}}_{\text{4}}$ and sodium hydroxide is,

$M_1=\text{1}\text{.00M}$ and $M_2\text{=}\text{0}\text{.220M}$

Calculation:

• The molarity formula is,

  $M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}$

And,

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

The balanced chemical equation is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\text{+2NaOH}\left(\text{aq}\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}$

Hence, 1 mole of Sulphuric acid is neutralizes the 2 moles of sodium hydroxide.

Firstly, calculate the number of moles of hydrogen ion in $0.220{\text{M H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution such that,

  $\begin{array}{l}{n}_{H^{+}}=M\times V=\left(\text{0}\text{.220}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\right)\text{×}\left(\text{35}{\text{.0×10}}^{\text{-3}}\text{L}\right)\\ n_{H^{+}}=\text{7}{\text{.7×10}}^{\text{-3}}\text{mol}\end{array}$

For neutralizing the reaction, the number of hydrogen ions must be equal to the number of hydroxide ions.

According to the balanced equations both hydrogen and hydroxide ions are reacted in a $1:2$ ratio.

Therefore, the number of moles of hydroxide ions are,

  $\text{0}\text{.0077moles×}\frac{\text{2}\text{moles}\text{NaOH}}{\text{1mole}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=0}\text{.0154moles}$

Now, calculate the volume of the $1.00MNaOH$ solution for 1 mole hydroxide ion is,

  $\begin{array}{l}n=M\times V\\ \Rightarrow V_{O{H}^{-}}=\frac{n}{M}\\ =\frac{\text{0}\text{.0154}\text{moles}}{\text{1}\text{.00}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}\text{=}\text{15}\text{.4mL}\end{array}$

The volume is $15.4mL$ .