Chapter 17, Problem 17.83SP

Interpretation Introduction

(a)

Interpretation:

$\text{pH}$ after the addition of $3.\text{0 mL}$ of $\text{KOH}$ 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 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 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

(b)

Interpretation:

$\text{pH}$ after the addition of $20\text{ mL}$ of $\text{KOH}$ 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:

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

Interpretation Introduction

(c)

Interpretation:

$\text{pH}$ after the addition of $65.0\text{ mL}$ of $\text{KOH}$ 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:

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