Chapter 17, Problem 17.85SP

Interpretation Introduction

(a)

Interpretation:

$\text{NaOH}$ required at the equivalence point is to be calculated and $\text{pH}$ after addition of $10\text{ mL}$ base is to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation is as follows:

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{base}\right]}{\left[\text{acid}\right]}$

The molarity is the concentration of the solution and is equal to the number of moles of the solute dissolved in a liter of the solution.

The formula to calculate molarity is given as follows:

$\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is as follows:

$1\text{ L}=\text{1000 mL}$

The negative logarithm of the molar concentration of hydronium ion is called $\text{pH}$. The expression for $\text{pH}$ is as follows:

$\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

The relation between ${\text{K}}_{\text{a}}$ and ${\text{pK}}_{\text{a}}$ is as follows:

${\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}$

Interpretation Introduction

(b)

Interpretation:

$\text{pH}$ at halfway to equivalence point is to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation is as follows:

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{base}\right]}{\left[\text{acid}\right]}$

The molarity is the concentration of the solution and is equal to the number of moles of the solute dissolved in a liter of the solution.

The formula to calculate molarity is given as follows:

$\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is as follows:

$1\text{ L}=\text{1000 mL}$

The negative logarithm of the molar concentration of hydronium ion is called $\text{pH}$. The expression for pH is as follows:

$\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

$\text{pH}+\text{pOH}=\text{14}$

The relation between ${\text{K}}_{\text{a}}$ and ${\text{pK}}_{\text{a}}$ is as follows:

${\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}$

Interpretation Introduction

(c)

Interpretation:

$\text{pH}$ at the equivalence point is to be calculated.

Concept introduction:

The molarity is the concentration of the solution and is equal to the number of moles of the solute dissolved in a liter of the solution.

The formula to calculate molarity is given as follows:

$\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is as follows:

$1\text{ L}=\text{1000 mL}$

The negative logarithm of the molar concentration of hydronium ion is called $\text{pH}$. The expression for pH is as follows:

$\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

$\text{pH}+\text{pOH}=\text{14}$

$K_{\text{a}}$ is defined as the acid dissociation constant in conjugate acid-base pairs.

$K_{\text{b}}$ is defined as the base ionization constant in acid-base pairs.

$K_{\text{w}}$ is the equilibrium constant for water. At $25°\text{C}$, $K_{\text{w}}$ is equal to $1.0\times {10}^{-14}$.

The expression for relation between $K_{\text{b}}$ and $K_{\text{a}}$ is as follows:

$K_{\text{a}}\cdot K_{\text{b}}=K_{\text{w}}$

At $25°\text{C}$, $K_{\text{w}}$ is equal to $1.0\times {10}^{-14}$, the above equation will be modified as follows:

$K_{\text{a}}\cdot K_{\text{b}}=1.0\times {10}^{-14}$

d)

Interpretation Introduction

Interpretation:

$\text{pH}$ after addition of $80\text{ mL}$ base is to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation is as follows:

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{base}\right]}{\left[\text{acid}\right]}$

The molarity is the concentration of the solution and is equal to the number of moles of the solute dissolved in a liter of the solution.

The formula to calculate molarity is given as follows:

$\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is as follows:

$1\text{ L}=\text{1000 mL}$

The negative logarithm of the molar concentration of hydronium ion is called $\text{pH}$. The expression for $\text{pH}$ is as follows:

$\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

The relation between ${\text{K}}_{\text{a}}$ and ${\text{pK}}_{\text{a}}$ is as follows:

${\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}$