Chapter 15, Problem 129AP

What volume of 0.250 M HCI is required to neutralize each of the following solutions?

a. 25.0 mL of 0.103 M sodium hydroxide, NaOH

b. 50.0 mL of 0.00501 M calcium hydroxide, Ca(OH)₂

c. 20.0 mL of 0.226 M ammonia, NH₃

d. 15.0 mL of 0.0991 M potassium hydroxide, KOH

Interpretation Introduction

(a)

Interpretation:

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $25.0\text{mL}$ of $0.103\text{M}\text{NaOH}$ is to be calculated.

Concept Introduction:

Neutralization is the process in which an acid and a 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 129AP

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $25.0\text{mL}$ of $0.103\text{M}\text{NaOH}$ is $10.3\text{mL}$.

Explanation of Solution

The molarity of the given $\text{HCl}$ solution is $0.250\text{M}$.

The molarity of the given $\text{NaOH}$ solution is $0.103\text{M}$.

The volume of the $0.103\text{M}\text{NaOH}$ is $25.0\text{mL}$.

The neutralization reaction between $\text{HCl}$ and $\text{NaOH}$ is represented as:

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

The relation between molarities of $\text{HCl}$ and $\text{NaOH}$ is given as:

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

Where,

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

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

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

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

$\begin{array}{c}{\text{V}}_1=\frac{\left(0.103\text{M}\right)\left(25.0\text{mL}\right)}{0.250\text{M}}\\ =10.3\text{mL}\end{array}$

Therefore, volume of $0.250\text{M}\text{HCl}$ required to neutralize $25.0\text{mL}$ of $0.103\text{M}\text{NaOH}$ is $10.3\text{mL}$.

Interpretation Introduction

(b)

Interpretation:

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $50.0\text{mL}$ of $0.00501\text{M}\text{Ca}{\left(\text{OH}\right)}_2$ is to be calculated.

Concept Introduction:

Neutralization is the process in which an acid and a 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 129AP

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $50.0\text{mL}$ of $0.00501\text{M}\text{Ca}{\left(\text{OH}\right)}_2$ is $2.004\text{mL}$.

Explanation of Solution

The molarity of the given $\text{HCl}$ solution is $0.250\text{M}$.

The molarity of the given $\text{Ca}{\left(\text{OH}\right)}_2$ solution is $0.00501\text{M}$.

The volume of the $0.00501\text{M}\text{Ca}{\left(\text{OH}\right)}_2$ is $50.0\text{mL}$.

The neutralization reaction between $\text{HCl}$ and $\text{Ca}{\left(\text{OH}\right)}_2$ is represented as:

$\text{2HCl}\left(\text{a}\text{q}\right)+\text{Ca}{\left(\text{OH}\right)}_2\left(\text{a}\text{q}\right)\to {\text{CaCl}}_2\left(\text{a}\text{q}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The relation between molarities of $\text{HCl}$ and $\text{Ca}{\left(\text{OH}\right)}_2$ is given as:

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

Where,

• ${\text{M}}_1$ represents the molarity of $\text{HCl}$.
• ${\text{V}}_1$ represents the volume of $\text{HCl}$.
• ${\text{M}}_2$ represents the molarity of $\text{Ca}{\left(\text{OH}\right)}_2$.
• ${\text{V}}_2$ represents the volume of $\text{Ca}{\left(\text{OH}\right)}_2$.

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

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

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

$\begin{array}{c}{\text{V}}_1=\frac{\left(2\right)\left(0.00501\text{M}\right)\left(50.0\text{mL}\right)}{0.250\text{M}}\\ =2.004\text{mL}\end{array}$

Therefore, volume of $0.250\text{M}\text{HCl}$ required to neutralize $50.0\text{mL}$ of $0.00501\text{M}\text{Ca}{\left(\text{OH}\right)}_2$ is $2.004\text{mL}$.

Interpretation Introduction

(c)

Interpretation:

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $20.0\text{mL}$ of $0.226\text{M}{\text{NH}}_3$ is to be calculated.

Concept Introduction:

Neutralization is the process in which an acid and a 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 129AP

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $20.0\text{mL}$ of $0.226\text{M}{\text{NH}}_3$ is $18.08\text{mL}$.

Explanation of Solution

The molarity of the given $\text{HCl}$ solution is $0.250\text{M}$.

The molarity of the given ${\text{NH}}_3$ solution is $0.226\text{M}$.

The volume of the $0.226\text{M}{\text{NH}}_3$ is $20.0\text{mL}$.

The neutralization reaction between $\text{HCl}$ and ${\text{NH}}_3$ is represented as:

$\text{HCl}\left(\text{a}\text{q}\right)+{\text{NH}}_3\left(\text{a}\text{q}\right)\to {\text{NH}}_4\text{Cl}\left(\text{a}\text{q}\right)$

The relation between molarities of $\text{HCl}$ and ${\text{NH}}_3$ is given as:

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

Where,

• ${\text{M}}_1$ represents the molarity of $\text{HCl}$.
• ${\text{V}}_1$ represents the volume of $\text{HCl}$.
• ${\text{M}}_2$ represents the molarity of ${\text{NH}}_3$.
• ${\text{V}}_2$ represents the volume of ${\text{NH}}_3$.

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

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

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

$\begin{array}{c}{\text{V}}_1=\frac{\left(0.226\text{M}\right)\left(20.0\text{mL}\right)}{0.250\text{M}}\\ =18.08\text{mL}\end{array}$

Therefore, volume of $0.226\text{M}{\text{NH}}_3$ required to neutralize $20.0\text{mL}$ of $0.226\text{M}{\text{NH}}_3$ is $18.08\text{mL}$.

Interpretation Introduction

(d)

Interpretation:

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $15.0\text{mL}$ of $0.0991\text{M}\text{KOH}$ is to be calculated.

Concept Introduction:

Neutralization is the process in which an acid and a 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 129AP

The volume of $0.250\text{M}\text{HCl}$ required to neutralize $15.0\text{mL}$ of $0.0991\text{M}\text{KOH}$ is $7.928\text{mL}$.

Explanation of Solution

The molarity of the given $\text{HCl}$ solution is $0.250\text{M}$.

The molarity of the given $\text{KOH}$ solution is $0.0991\text{M}$.

The volume of the $0.0991\text{M}\text{KOH}$ is $15.0\text{mL}$.

The neutralization reaction between $\text{HCl}$ and $\text{KOH}$ is represented as:

$\text{HCl}\left(\text{a}\text{q}\right)+\text{KOH}\left(\text{a}\text{q}\right)\to \text{KCl}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The relation between molarities of $\text{HCl}$ and $\text{KOH}$ is given as:

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

Where,

• ${\text{M}}_1$ represents the molarity of $\text{HCl}$.
• ${\text{V}}_1$ represents the volume of $\text{HCl}$.
• ${\text{M}}_2$ represents the molarity of $\text{KOH}$.
• ${\text{V}}_2$ represents the volume of $\text{KOH}$.

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

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

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

$\begin{array}{c}{\text{V}}_1=\frac{\left(0.0991\text{M}\right)\left(20.0\text{mL}\right)}{0.250\text{M}}\\ =7.928\text{mL}\end{array}$

Therefore, volume of $0.250\text{M}\text{HCl}$ required to neutralize $15.0\text{mL}$ of $0.0991\text{M}\text{KOH}$ is $7.928\text{mL}$.