Chapter 17, Problem 17.88SP

Interpretation Introduction

(a)

Interpretation:

The $\text{pH}$ titration curve should be sketched. Also, the equivalence point is to be labeled.

Concept introduction:

Titration is a process that determines the concentration of a solution of known volume that reacts with a standard solution of other substances.

$\text{pH}$ titration curve is plot of pH of the solution versus volume of titrant. The shape of $\text{pH}$ the titration curve identifies the equivalence point in the titration.

The point at which an equal quantity of acid and base have mixed together is called equivalence point. $\text{pH}$ at equivalence point depends upon the relative strength of acids or bases.

Interpretation Introduction

(b)

Interpretation:

$\text{NaOH}$ required to reach 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 solute dissolved per 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:

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

Interpretation Introduction

(c)

Interpretation:

The $\text{pH}$ at the equivalence point is to be calculated.Whether $\text{pH}$ at the equivalence point is greater than, equal to, or less than 7 should be determined.

Concept introduction:

The molarity is the concentration of the solution and is equal to the number of moles of solute dissolved per 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 equation (1) will be modified as follows:

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

d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ at exactly halfway to the 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 solute dissolved per 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:

${\text{pK}}_{\text{a}}=-\log \left({\text{K}}_{\text{a}}\right)$