Chapter 15, Problem 45QRT

(a)

Interpretation Introduction

Interpretation:

pH of $\text{50mL}\text{of}\text{0}\text{.150}\text{M}\text{NaOH}$ solution has to be calculated.

Concept Introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

$\text{pOH}$ Definition:

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

  $\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Explanation of Solution

Given,

Volume of acid in liters:

  $\text{=50}\text{mL}\text{×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{=0}\text{.05000L}$

Original mole ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$added:

  $\text{=0}\text{.05000L}\text{×}\frac{\text{0}\text{.150}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}}{\text{1L}}\text{=0}\text{.00750}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

The given solution contains only$\text{0}\text{.150 M}\text{HCl}$, which ionizes to make $\text{0}\text{.150 M}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

  $\begin{array}{c}\text{pH=log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{=-log(0}\text{.150)}\\ \text{=}\text{0}\text{.824}\end{array}$

(b)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of $\text{25}\text{.0}\text{mL}\text{of }\text{0}\text{.150}\text{M}\text{HCl}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

After the addition of the titrant,

To determine the equilibrium concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{=}\frac{\text{original}\text{moles}\text{acid-total}\text{moles}\text{base}\text{added}}{\text{volume}\text{of}\text{acid}\text{(L)+volume}\text{of}\text{base}\text{(L)}}\\ \\ \text{volume}\text{of}\text{base}\text{in}\text{liters}\text{=25}\text{.00}\text{mL}\text{×}\frac{\text{1L}}{\text{1000}\text{mL}}\text{=0}\text{.02500}\text{L}\\ \\ \text{total}\text{mol}{\text{OH}}^{\text{-}}\text{added}\text{=}\text{0}\text{.02500}\text{mL}\text{×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\text{=0}\text{.00375L}\text{mol}{\text{OH}}^{\text{-}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{0}\text{.00750}\text{mol}\text{-0}\text{.00375}\text{mol}}{\text{0}\text{.05000L+0}\text{.02500}\text{L}}\text{=}\text{0}\text{.0500}\text{M}\\ \\ \text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{=-log(0}\text{.050)}\\ \text{=}\text{1}\text{.30}\end{array}$

(c)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of$\text{49}\text{.9 mL}\text{of }\text{0}\text{.150}\text{M}\text{HCl}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Used the equation in (b), again

Volume of base (L)

  $\begin{array}{l}\text{49}\text{.9}\text{mL}\text{×}\frac{\text{1L}}{\text{1000mL}}\text{=0}\text{.0490L}\\ \\ \text{total}\text{mol}{\text{OH}}^{\text{-}}\text{added}\text{=}\text{0}\text{.0499}\text{L}\text{×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\text{=}\text{0}\text{.007485}\text{mol}{\text{OH}}^{\text{-}}\\ \\ \text{=}\text{0}\text{.00749}\text{mol}{\text{OH}}^{\text{-}}\end{array}$

  $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{0}\text{.00750}\text{mol}\text{-0}\text{.007485}\text{mol}}{\text{0}\text{.05000L+0}\text{.0499}\text{L}}\\ \\ \text{=}\frac{\text{0}\text{.000015}\text{mol}}{\text{0}\text{.0999}\text{L}}\\ \\ \text{=0}\text{.00015M}\approx \text{0}\text{.0002M}\\ \\ \text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{=-log(0}\text{.00015)}\\ \approx \text{3}\text{.8}\text{.}\end{array}$

(d)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of$\text{50}\text{.0}\text{mL}\text{of }\text{0}\text{.150}\text{M}\text{HCl}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

The volume and total moles of base can be calculated as

  $\begin{array}{l}\text{volume}\text{of}\text{base}\text{(L)}\text{=50}\text{mL}\text{×}\frac{\text{1L}}{\text{1000mL}}\text{=0}\text{.05000L}\\ \\ \text{total}\text{mol}{\text{OH}}^{\text{-}}\text{added}\text{=}\text{0}\text{.05000}\text{L}\text{×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\text{=}\text{0}\text{.00750}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

Total moles of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$added is equal to the original moles ${\text{OH}}^{\text{-}},$so the solution is neutral, and the $\text{pH = 7}\text{.000}\text{.}$

(e)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of $\text{50}\text{.1mL}\text{of }\text{0}\text{.150}\text{M}\text{HCl}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

At equivalent point, the moles of base begin to exceed the moles of acid, so we adapt the expression we used in (b) for a solution with excess base to determine the equilibrium concentration of ${\text{OH}}^{\text{–}}$.

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{total}\text{moles}\text{base}\text{added}\text{-}\text{original}\text{moles}\text{acid}}{\text{volume}\text{of}\text{acid(L)}\text{+}\text{volume}\text{of}\text{base}\text{(L)}}\\ \\ \text{volume}\text{of}\text{base}\text{(L)}\text{=50}\text{.1mL}\text{×}\frac{\text{1L}}{\text{1000mL}}\text{=0}\text{.0501L}\\ \\ \text{total}\text{mol}{\text{OH}}^{\text{-}}\text{added}\text{=}\text{0}\text{.0501}\text{L}\text{×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{G}}}{\text{1L}}\text{=}\text{0}\text{.00750}\text{mol}{\text{OH}}^{\text{G}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{0}\text{.007515}\text{mol}\text{-0}\text{.00750}\text{mol}}{\text{0}\text{.05000L+0}\text{.0501}\text{L}}\text{=}\frac{\text{0}\text{.000015}\text{mol}}{\text{0}\text{.1001}\text{L}}\\ \\ \text{=0}\text{.00015M}\\ \\ \text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{=-log(0}\text{.000015)}\\ \text{=3}\text{.8}\\ \text{pH=14}\text{.00-pOH}\\ \text{=}\text{14}\text{.00-3}\text{.8=10}\text{.2}\end{array}$

The solution has a $\text{pH}$ of $10.2$.

(f)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of$\text{75}\text{.0}\text{mL}\text{of }\text{0}\text{.150}\text{M}\text{HCl}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

The pH of equivalent point of titrant can be calculated as

$\begin{array}{l}\text{volume}\text{of}\text{base}\text{(L)}\text{=75}\text{mL}\text{×}\frac{\text{1L}}{\text{1000mL}}\text{=0}\text{.0750L}\\ \\ \text{total}\text{mol}{\text{OH}}^{\text{-}}\text{added}\text{=}\text{0}\text{.0750}\text{L}\text{×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\text{=}\text{0}\text{.0113}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

  $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{0}\text{.0113}\text{mol}\text{-0}\text{.00750}\text{mol}}{\text{0}\text{.05000L+0}\text{.0750}\text{L}}\\ \\ \text{=}\frac{\text{0}\text{.00038}\text{mol}}{\text{0}\text{.1250}\text{L}}\\ \\ \text{=0}\text{.030M}\\ \\ \text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{=-log(0}\text{.030)}\\ \text{=1}\text{.52}\\ \text{pH=14}\text{.00-pOH}\\ \text{=}\text{14}\text{.00-1}\text{.52=12}\text{.48}\end{array}$

Hence, the results in a pH verses volume of titrant to make a titration curve, including equivalence point: