Chapter 9, Problem 9.99E

Interpretation Introduction

(a)

Interpretation:

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is to be determined.

Concept introduction:

Molarity is the ratio of the number of moles of solute to the volume of the solution in liters.

The molarity is given by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$

Answer to Problem 9.99E

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $33.33\text{mL}$.

Explanation of Solution

The volume and molarity of $\text{HCl}$ is $\text{20}\text{.00}\text{mL}$ and $\text{0}\text{.200}\text{M}$ respectively.

The number of millimoles of $\text{HCl}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$ …(1)

The above formula can be written as follows:

$\text{Number}\text{of}\text{millimoles}=\text{Molarity}\times \text{Volume}\left(\text{mL}\right)$ …(2)

Substitute the volume and molarity in equation (2).

$\begin{array}{rll}\text{Number}\text{of}\text{millimoles}& =& \text{0}\text{.200}\text{M}\times \text{20}\text{.00}\text{mL}\\ \text{Number}\text{of}\text{millimoles}& =& 4\text{mmol}\end{array}$

Thus, the number of millimoles of $\text{HCl}$ is $4\text{mmol}$.

The neutralization reaction is given below.

$\text{HCl}\left(\text{aq}\right)+\text{NaOH}\left(\text{aq}\right)\to \text{NaCl}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From the above equation, the molar ratio of $\text{HCl}$ and $\text{NaOH}$ is $1:1$. Hence, the number of moles of $\text{NaOH}$ is $4\text{mmol}$.

The given molarity of $\text{NaOH}$ is $\text{0}\text{.120}\text{M}$.

Substitute the molarity and millimoles of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{0}\text{.120}\text{M}& =& \frac{4\text{mmol}}{\text{Volume}\left(\text{mL}\right)}\\ \text{Volume}\left(\text{mL}\right)& =& \frac{4\text{mmol}}{0.120\text{M}}\\ & =& 33.33\text{mL}\end{array}$

Hence, the number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $33.33\text{mL}$.

Conclusion

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $33.33\text{mL}$.

Interpretation Introduction

(b)

Interpretation:

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is to be determined.

Concept introduction:

Molarity is the ratio of the number of moles of solute to the volume of the solution in liters.

The molarity is given by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$

Answer to Problem 9.99E

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $66.67\text{mL}$.

Explanation of Solution

The volume and molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{20}\text{.00}\text{mL}$ and $\text{0}\text{.200}\text{M}$ respectively.

The number of millimoles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$ …(1)

The above formula can be written as follows:

$\text{Number}\text{of}\text{millimoles}=\text{Molarity}\times \text{Volume}\left(\text{mL}\right)$ …(2)

Substitute the volume and molarity in equation (2).

$\begin{array}{rll}\text{Number}\text{of}\text{millimoles}& =& \text{0}\text{.200}\text{M}\times \text{20}\text{.00}\text{mL}\\ \text{Number}\text{of}\text{millimoles}& =& 4\text{mmol}\end{array}$

Thus, the number of millimoles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $4\text{mmol}$.

The neutralization reaction is given below.

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+2\text{NaOH}\left(\text{aq}\right)\to {\text{Na}}_2{\text{SO}}_{\text{4}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From the above equation, the molar ratio of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{NaOH}$ is $1:2$. Hence, the number of millimoles of $\text{NaOH}$ required to neutralize $4\text{mmol}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as follows:

$\begin{array}{rll}\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}& =& 2\times \text{mole}\text{of}\text{NaOH}\\ 4\text{mmol}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}& =& 2\times 4\text{mmol}\text{of}\text{NaOH}\\ & =& 8\text{mmol}\text{of}\text{NaOH}\end{array}$

Hence, the number of millimoles of $\text{NaOH}$ is $8\text{mmol}$.

The given molarity of $\text{NaOH}$ is $\text{0}\text{.120}\text{M}$.

Substitute the molarity and millimoles of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{0}\text{.120}\text{M}& =& \frac{8\text{mmol}}{\text{Volume}\left(\text{mL}\right)}\\ \text{Volume}\left(\text{mL}\right)& =& \frac{8\text{mmol}}{0.120\text{M}}\\ & =& 66.67\text{mL}\end{array}$

Hence, the number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $66.67\text{mL}$.

Conclusion

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $66.67\text{mL}$.

Interpretation Introduction

(c)

Interpretation:

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is to be determined.

Concept introduction:

Molarity is the ratio of the number of moles of solute to the volume of the solution in liters.

The molarity is given by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$

Answer to Problem 9.99E

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $41.67\text{mL}$.

Explanation of Solution

The volume and molarity of $\text{HCl}$ is $\text{20}\text{.00}\text{mL}$ and $\text{0}\text{.250}\text{M}$ respectively.

The number of millimoles of $\text{HCl}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$ …(1)

The above formula can be written as follows:

$\text{Number}\text{of}\text{millimoles}=\text{Molarity}\times \text{Volume}\left(\text{mL}\right)$ …(2)

Substitute the volume and molarity in equation (2).

$\begin{array}{rll}\text{Number}\text{of}\text{millimoles}& =& \text{0}\text{.250}\text{M}\times \text{20}\text{.00}\text{mL}\\ \text{Number}\text{of}\text{millimoles}& =& 5\text{mmol}\end{array}$

Thus, the number of millimoles of $\text{HCl}$ is $5\text{mmol}$.

The neutralization reaction is given below.

$\text{HCl}\left(\text{aq}\right)+\text{NaOH}\left(\text{aq}\right)\to \text{NaCl}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From the above equation, the molar ratio of $\text{HCl}$ and $\text{NaOH}$ is $1:1$. Hence, the number of millimoles of $\text{NaOH}$ is $5\text{mmol}$.

The given molarity of $\text{NaOH}$ is $\text{0}\text{.120}\text{M}$.

Substitute the molarity and moles of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{0}\text{.120}\text{M}& =& \frac{5\text{mmol}}{\text{Volume}\left(\text{mL}\right)}\\ \text{Volume}\left(\text{mL}\right)& =& \frac{5\text{mmol}}{0.120\text{M}}\\ & =& 41.67\text{mL}\end{array}$

Hence, the number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $41.67\text{mL}$.

Conclusion

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $41.67\text{mL}$.

Interpretation Introduction

(d)

Interpretation:

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is to be determined.

Concept introduction:

Molarity is the ratio of the number of moles of solute to the volume of the solution in liters.

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Answer to Problem 9.99E

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $340.1\text{mL}$.

Explanation of Solution

The number of moles a substance is given as,

$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance

The number of moles of $\text{HCl}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$ …(2)

The above formula can be written as follows:

$\text{Number}\text{of}\text{moles}=\frac{\text{Molarity}\times \text{Volume}\left(\text{mL}\right)\times 1\text{L}}{1000\text{mL}}$ …(3)

Equate equation (1) and (3).

$\frac{m}{M}=\frac{\text{Molarity}\times \text{Volume}\left(\text{mL}\right)\times 1\text{L}}{1000\text{mL}}$ …(4)

The molar mass of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $98\text{g}{\text{mol}}^{-1}$. The given mass and volume of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{10}\text{.00}\text{g}$ and $\text{150}\text{.}\text{mL}$ respectively.

Substitute the molar mass and given mass of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ in equation (4).

$\begin{array}{rll}\frac{10\text{g}}{98\text{g}{\text{mol}}^{-1}}& =& \frac{\text{Molarity}\times \text{150}\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Molarity}& =& \frac{10\text{g}\times 1000\text{mL}}{98\text{g}{\text{mol}}^{-1}\times \text{150}\text{mL}\times 1\text{L}}\\ & =& 0.6802\text{M}\end{array}$

Thus, the molarity of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $0.6802\text{M}$. The given volume of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{20}\text{.00}\text{mL}$.

The number of millimoles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$ …(5)

The above formula can be written as follows:

$\text{Number}\text{of}\text{millimoles}=\text{Molarity}\times \text{Volume}\left(\text{mL}\right)$

Substitute the volume and molarity in above formula.

$\begin{array}{rll}\text{Number}\text{of}\text{millimoles}& =& \text{0}\text{.6802}\text{M}\times \text{20}\text{.00}\text{mL}\\ \text{Number}\text{of}\text{millimoles}& =& 13.604\text{mmol}\end{array}$

Thus, the number of millimoles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $13.604\text{mmol}$.

The neutralization reaction is given below.

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

From the above equation, the molar ratio of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and $\text{NaOH}$ is $1:3$. Hence, the number of millimoles of $\text{NaOH}$ required to neutralize $13.604\text{mmol}$ of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated as follows:

$\begin{array}{rll}\text{1}\text{mole}\text{of}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}& =& 3\times \text{mole}\text{of}\text{NaOH}\\ 13.604\text{mmol}\text{of}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}& =& 3\times 13.604\text{mmol}\text{of}\text{NaOH}\\ & =& 40.812\text{mmole}\text{of}\text{NaOH}\end{array}$

Hence, the number of millimoles of $\text{NaOH}$ is $40.812\text{mmol}$.

The given molarity of $\text{NaOH}$ is $\text{0}\text{.120}\text{M}$.

Substitute the molarity and moles of $\text{NaOH}$ in equation (5).

$\begin{array}{rll}\text{0}\text{.120}\text{M}& =& \frac{40.812\text{mmol}}{\text{Volume}\left(\text{mL}\right)}\\ \text{Volume}\left(\text{mL}\right)& =& \frac{40.812\text{mmol}}{0.120\text{M}}\\ & =& 340.1\text{mL}\end{array}$

Hence, the number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $340.1\text{mL}$.

Conclusion

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $340.1\text{mL}$.

Interpretation Introduction

(e)

Interpretation:

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is to be determined.

Concept introduction:

Molarity is the ratio of the number of moles of solute to the volume of the solution in liters.

The molarity is given by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Answer to Problem 9.99E

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $125\text{mL}$.

Explanation of Solution

The moles and volume of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is $\text{0}\text{.150}\text{mol}$ and $\text{400}\text{.}\text{mL}$ respectively.

The molarity of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Substitute the volume and moles in above formula.

$\begin{array}{rll}\text{Molarity}& =& \frac{\text{0}\text{.150}\text{mol}}{\text{400}\text{mL}}\times \frac{1000\text{mL}}{1\text{L}}\\ & =& 0.375\text{M}\end{array}$

Thus, the molarity of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is $0.375\text{M}$.

The number of millimoles of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$ …(1)

The above formula can be written as follows:

$\text{Number}\text{of}\text{millimoles}=\text{Molarity}\times \text{Volume}\left(\text{mL}\right)$

Substitute the volume and molarity in above formula.

$\begin{array}{rll}\text{Number}\text{of}\text{millimoles}& =& 0.375\text{M}\times \text{20}\text{.00}\text{mL}\\ \text{Number}\text{of}\text{millimoles}& =& 7.5\text{mmol}\end{array}$

Thus, the number of millimoles of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is $7.5\text{mmol}$.

The neutralization reaction is given below.

${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}\left(\text{aq}\right)+2\text{NaOH}\left(\text{aq}\right)\to {\text{Na}}_2{\text{MoO}}_{\text{4}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From the above equation, the molar ratio of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ and $\text{NaOH}$ is $1:2$. Hence, the number of millimoles of $\text{NaOH}$ required to neutralize $7.5\text{mmol}$ of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is calculated as follows:

$\begin{array}{rll}\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}& =& 2\times \text{mole}\text{of}\text{NaOH}\\ 7.5\text{mmol}\text{of}{\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}& =& 2\times 7.5\text{mmol}\text{of}\text{NaOH}\\ & =& 15\text{mmole}\text{of}\text{NaOH}\end{array}$

Hence, the number of millimoles of $\text{NaOH}$ is $15\text{mmol}$.

The given molarity of $\text{NaOH}$ is $\text{0}\text{.120}\text{M}$.

Substitute the molarity and moles of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{0}\text{.120}\text{M}& =& \frac{15\text{mmol}}{\text{Volume}\left(\text{mL}\right)}\\ \text{Volume}\left(\text{mL}\right)& =& \frac{15\text{mmol}}{0.120\text{M}}\\ & =& 125\text{mL}\end{array}$

Hence, the number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $125\text{mL}$.

Conclusion

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $125\text{mL}$.

Interpretation Introduction

(f)

Interpretation:

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is to be determined.

Concept introduction:

Molarity is the ratio of the number of moles of solute to the volume of the solution in liters.

The molarity is given by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Answer to Problem 9.99E

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $119.3\text{mL}$.

Explanation of Solution

The moles and volume of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is $\text{0}\text{.215}\text{mol}$ and $\text{600}\text{.}\text{mL}$ respectively.

The molarity of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Substitute the volume and moles in above formula.

$\begin{array}{c}\text{Molarity}=\frac{\text{0}\text{.215}\text{mol}}{\text{600}\text{mL}}\times \frac{1000\text{mL}}{1\text{L}}\\ =0.358\text{M}\end{array}$

Thus, the molarity of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is $0.358\text{M}$.

The number of millimoles of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{millimoles}}{\text{Volume}\left(\text{mL}\right)}$ …(1)

The above formula can be written as follows:

$\text{Number}\text{of}\text{millimoles}=\text{Molarity}\times \text{Volume}\left(\text{mL}\right)$

Substitute the volume and molarity in above formula.

$\begin{array}{rll}\text{Number}\text{of}\text{millimoles}& =& 0.358\text{M}\times \text{20}\text{.00}\text{mL}\\ \text{Number}\text{of}\text{millimoles}& =& 7.16\text{mmol}\end{array}$

Thus, the number of millimoles of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is $7.16\text{mmol}$.

The neutralization reaction is given below.

${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}\left(\text{aq}\right)+2\text{NaOH}\left(\text{aq}\right)\to {\text{Na}}_2{\text{MoO}}_{\text{4}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From the above equation, the molar ratio of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ and $\text{NaOH}$ is $1:2$. Hence, the number of millimoles of $\text{NaOH}$ required to neutralize $7.5\text{mmol}$ of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ is calculated as follows:

$\begin{array}{c}\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}=2\times \text{mole}\text{of}\text{NaOH}\\ 7.16\text{mmol}\text{of}{\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}=2\times 7.16\text{mmol}\text{of}\text{NaOH}\\ =14.32\text{mmole}\text{of}\text{NaOH}\end{array}$

Hence, the number of millimoles of $\text{NaOH}$ is $14.32\text{mmol}$.

The given molarity of $\text{NaOH}$ is $\text{0}\text{.120}\text{M}$.

Substitute the molarity and moles of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{0}\text{.120}\text{M}& =& \frac{14.32\text{mmol}}{\text{Volume}\left(\text{mL}\right)}\\ \text{Volume}\left(\text{mL}\right)& =& \frac{14.32\text{mmol}}{0.120\text{M}}\\ & =& 119.3\text{mL}\end{array}$

Hence, the number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $119.3\text{mL}$.

Conclusion

The number of milliliters of $\text{NaOH}$ solution that will be needed for given acid sample is $119.3\text{mL}$.