Chapter 17, Problem 17.97SP

Interpretation Introduction

(a)

Interpretation:

$\text{pH}$ after addition of $25\text{ mL}$ base is to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation is as follows:

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

The molarity is concentration of solution and is equal to number of moles of solute dissolved per liter of 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 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 $K_{\text{a}}$ and $\text{p}{K}_{\text{a}}$ is as follows:

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Interpretation Introduction

(b)

Interpretation:

$\text{pH}$ after addition of $50\text{ mL}$ base is to be calculated.

Concept introduction:

The molarity is concentration of solution and is equal to number of moles of solute dissolved in liter of 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 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 $K_{\text{a}}$ and $\text{p}{K}_{\text{a}}$ is as follows:

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Interpretation Introduction

(c)

Interpretation:

$\text{pH}$ after addition of $75\text{ mL}$ base is to be calculated.

Concept introduction:

The molarity is concentration of the solution and is equal to number of moles of solute dissolved in liter of 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 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 $K_{\text{a}}$ and $\text{p}{K}_{\text{a}}$ is as follows:

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

d)

Interpretation Introduction

Interpretation:

$\text{pH}$ after addition of $100\text{ mL}$ base is to be calculated.

Concept introduction:

The molarity is concentration of the solution and is equal to number of moles ofsolute dissolved in liter of 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 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:

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