Chapter 9, Problem 9.105QP

(a)

Interpretation Introduction

Interpretation:

The required volume of $\text{NaOH}$ solution to neutralize the given acid solutions should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Answer to Problem 9.105QP

• The required volume of $\text{NaOH}$ solution to neutralize the given $\text{HCl}$ solution is $\text{59}\text{.85}\text{mL}\text{.}$

Explanation of Solution

Record the given data,

Concentration of $\text{NaOH}$= $\text{1}\text{.015 M}$

Volume of $\text{HCl}$ solution = $\text{25}\text{.00 mL}$

Concentration of $\text{HCl}$ solution = $\text{2}\text{.430 M}$

The required mole of $\text{NaOH}$to neutralize the given $\text{HCl}$solution

The balanced equation is,

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

$\begin{array}{l}\text{NaOH}\text{mole}\text{=}\text{25}\text{.0mL×}\frac{\text{2}\text{.430}\text{mol}\text{HCl}}{\text{1000}\text{mL}\text{soln}}\text{×}\frac{\text{1}\text{molNaOH}}{\text{1}\text{mol}\text{HCl}}\\ \text{=}\text{6}\text{.075}{\text{×10}}^{\text{-2}}\text{mol}\end{array}$

• $\text{HCl}$ is required 1:1 mole to neutralize the $\text{NaOH}$ so the volume and concentration of $\text{HCl}$ are plugged in the equation to give the required mole of $\text{NaOH}$ to neutralize the given $\text{HCl}$ solution.
• The required mole of $\text{HCl}$ to neutralize the given $\text{NaOH}$ solution $\mathrm{HCl}$ is $\text{6}\text{.075}{\text{×10}}^{\text{-2}}\text{mol}\text{.}$

The required volume of $\text{NaOH}$ to neutralize the given $\text{HCl}$ solution

$\begin{array}{l}\text{Volume}\text{of}\text{NaOH}\text{=}\frac{\text{6}{\text{.075×10}}^{\text{-2}}\text{molNaOH}}{\text{1}\text{.015}\text{mol/L}}\\ \text{=}\text{5}\text{.985}{\text{×10}}^{\text{-2}}\text{L}\\ \text{=}\text{59}\text{.85}\text{mL}\end{array}$

• The calculated required mol of $\text{NaOH}$ is divide by the concentrations of the $\text{NaOH}$ solution to give the required volume of $\text{NaOH}$ solution to neutralize the given $\text{HCl}$ solution.
• The required volume of $\text{NaOH}$ solution to neutralize the given $\text{HCl}$ solution is $\text{59}\text{.85}\text{mL}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The required volume of $\text{NaOH}$ solution to neutralize the given acid solutions should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Answer to Problem 9.105QP

• The required volume of $\text{NaOH}$ solution to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $\text{221}\text{.7}\text{mL}$

Explanation of Solution

Record the given data,

Concentration of $\text{NaOH}$= $\text{1}\text{.015 M}$

Volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution = $\text{25}\text{.00 mL}$

Concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution = $\text{4}\text{.500 M}$

The required mole of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution

The balanced equation is,

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

$\begin{array}{l}\text{NaOH}\text{mole}\text{=}\text{25}\text{.0}\text{mL×}\frac{\text{4}\text{.500 mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1000}\text{mL}\text{soln}}\text{×}\frac{\text{2}\text{mol}\text{NaOH}}{\text{1}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}\text{0}\text{.2250}\text{mol}\end{array}$

• The ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is required 1:2 mole to neutralize $\text{NaOH}$ so the volume and concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are plugged in the equation to give the required mole of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution $\text{NaOH}$
• The required mole of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $\text{0}\text{.2250}\text{mol}\text{.}$

The required volume of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution

$\begin{array}{l}\text{Volume}\text{of}\text{NaOH}\text{=}\frac{\text{2}{\text{.250×10}}^{\text{-1}}\text{mol}\text{NaOH}}{\text{1}\text{.015}\text{mol/L}}\\ \text{=}\text{2}{\text{.217×10}}^{\text{-1}}\text{L}\\ \text{=}\text{221}\text{.7}\text{mL}\end{array}$

• The calculated required mol of $\text{NaOH}$ is divide by the concentrations of the $\text{NaOH}$ solution to give the required volume of $\text{NaOH}$ solution to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.
• The required volume of $\text{NaOH}$ solution to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $\text{221}\text{.7}\text{mL}$

(c)

Interpretation Introduction

Interpretation:

The required volume of $\text{NaOH}$ solution to neutralize the given acid solutions should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Answer to Problem 9.105QP

• The required volume of $\text{NaOH}$ solution to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{221}\text{.7}\text{mL}$

Explanation of Solution

Record the given data,

Concentration of $\text{NaOH}$= $\text{1}\text{.015 M}$

Volume of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution = $\text{25}\text{.00 mL}$

Concentration of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution = $\text{1}\text{.500 M}$

The required mole of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution

The balanced equation is,

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

$\begin{array}{l}\text{NaOH}\text{mole}\text{=}\text{25}\text{.0}\text{mL×}\frac{\text{1}\text{.500 mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1000}\text{mL}\text{soln}}\text{×}\frac{3\text{mol}\text{NaOH}}{\text{1}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\\ \text{=}\text{0}\text{.1125}\text{mol}\end{array}$

• The ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is required 1:3 mole to neutralize $\text{NaOH}$ so the volume and concentration of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ are plugged in the equation to give the required mole of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution $\text{NaOH}$
• The required mole of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.1125}\text{mol}\text{.}$

The required volume of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution

$\begin{array}{l}\text{Volume}\text{of}\text{NaOH}\text{=}\frac{\text{1}{\text{.150×10}}^{\text{-1}}\text{mol}\text{NaOH}}{\text{1}\text{.015}\text{mol/L}}\\ \text{=}\text{1}\text{.108}{\text{×10}}^{\text{-1}}\text{L}\\ \text{=}\text{110}\text{.8}\text{mL}\end{array}$

• The calculated required mol of $\text{NaOH}$ is divided by the concentrations of the $\text{NaOH}$ solution to give the required volume of $\text{NaOH}$ solution to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution.
• The required volume of $\text{NaOH}$ solution to neutralize the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{221}\text{.7}\text{mL}$