Chapter 15, Problem 44QRT

(a)

Interpretation Introduction

Interpretation:

After volume of $\text{0 mL}$ titrant has been added, the $\text{pH}$ of the solution has to be calculated.

Concept Introduction:

The relationship between ${\text{pK}}_{\text{a}}{\text{ and pK}}_{\text{b}}$is given below,

  ${\text{pK}}_{\text{a}}{\text{ + pK}}_{\text{b}}\text{ =14}\text{.00 }$

The ${\text{K}}_a$ and ${\text{K}}_{\text{b}}$ value is calculated by using following formula,

  ${\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}$

The $\text{pH}$ and $\text{pOH}$ is expressed as follows,

  $\text{pH = -log}\left({\text{H}}^{\text{+}}\right)\text{ and}\text{pOH = -log}\left({\text{OH}}^{\text{-}}\right)$

The relationship between $\text{pH and pOH }$is given below,

  $\text{pH + pOH =14}\text{.00 }$

The concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$can be calculated by using following method,

  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{= }\frac{\text{Initial}\text{amount}\text{acid}\text{(mol)}\text{-}\text{Amount}\text{base}\text{added}\text{(mol)}}{\text{Initial}\text{volume acid (L)}\text{+}\text{Volume base}\text{added}\text{(L)}}$

The concentration of $\left[{\text{OH}}^{-}\right]$ can be calculated by using following method,

  $\left[{\text{OH}}^{-}\right]\text{= }\frac{\text{Initial}\text{amount}\text{base}\text{(mol)}\text{-}\text{Amount}\text{acid}\text{added}\text{(mol)}}{\text{Initial}\text{volume acid (L)}\text{+}\text{Volume base}\text{added}\text{(L)}}$

Explanation of Solution

Given,

The titration of $\text{50 mL}$ of $\text{0}\text{.150 M HCl}$ solution with $\text{0}\text{.150 M NaOH}$ solution.

Volume of the solution is given below,

  $\begin{array}{l}\text{Total volume of the solution in liters = 50}\text{.00 mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{Total volume of the solution in liters =}\text{0}\text{.05}\text{L}\end{array}$

Mole of hydroxide ion is calculated as follows,

  $\begin{array}{l}\text{Mole of hydroxide ion}\text{added = }\text{0}\text{.05}\text{L×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\\ \text{Mole of hydroxide ion}\text{added = 7}{\text{.5×10}}^{\text{-3}}\text{mol}{\text{OH}}^{\text{-}}\end{array}$

The $\text{pOH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pOH = -log}\left({\text{OH}}^{\text{-}}\right)\\ \text{pOH = -log}\left(0.150\right)\\ \text{pOH = 0}\text{.824}\end{array}$

The $\text{pH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pH + pOH =14}\text{.00}\\ \text{pH + 0}\text{.824 =14}\text{.00}\\ \text{pH =13}\text{.176 }\end{array}$

The $\text{pH}$ of the solution is calculated

(b)

Interpretation Introduction

Interpretation:

After volume of $\text{25 mL}$ titrant has been added, the $\text{pH}$ of the solution has to be calculated.

Concept Introduction:

Refer to part (a)

Explanation of Solution

Given,

The titration of $\text{50 mL}$ of $\text{0}\text{.150 M HCl}$ solution with $\text{0}\text{.150 M NaOH}$ solution.

Volume of the acid solution is given below,

  $\begin{array}{l}\text{Total volume of the acid solution in liters = 25}\text{.0 mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{Total volume of the solution in liters =}\text{0}\text{.025}\text{L}\end{array}$

Mole of hydronium ion is calculated as follows,

  $\begin{array}{l}{\text{Moles of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{added = }\text{0}\text{.025}\text{L×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\\ {\text{Mole of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{ = 3}{\text{.75×10}}^{\text{-3}}\text{mol}{\text{hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$is calculated as follows,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{Initial}\text{amount}\text{base}\text{(mol)}\text{-}\text{Amount}\text{acid}\text{added}\text{(mol)}}{\text{Initial}\text{volume acid (L)}\text{+}\text{Volume base}\text{added}\text{(L)}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{0}\text{.00750}\text{mol}\text{-}\text{0}\text{.00375}\text{mol}}{\text{0}\text{.05}\text{L}\text{+}\text{0}\text{.0250}\text{L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{{\text{4×10}}^{\text{-3}}\text{mol}}{\text{0}\text{.075L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= 0}\text{.0533 M}\end{array}$

The $\text{pOH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pOH = -log}\left({\text{OH}}^{\text{-}}\right)\\ \text{pOH = -log}\left(0.0533\right)\\ \text{pOH = 1}\text{.33}\end{array}$

The $\text{pH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pH + pOH =14}\text{.00}\\ \text{pH + 1}\text{.33 =14}\text{.00}\\ \text{pH =12}\text{.67 }\end{array}$

The $\text{pH}$ of the solution is calculated

(c)

Interpretation Introduction

Interpretation:

After volume of $\text{49}\text{.9 mL}$ titrant has been added, the $\text{pH}$ of the solution has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given,

The titration of $\text{50 mL}$ of $\text{0}\text{.150 M HCl}$ solution with $\text{0}\text{.150 M NaOH}$ solution.

Volume of the acid solution is given below,

  $\begin{array}{l}\text{Total volume of the acid solution in liters =49}\text{.9 mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{Total volume of the solution in liters =}\text{0}\text{.0499}\text{L}\end{array}$

Mole of hydronium ion is calculated as follows,

  $\begin{array}{l}{\text{Moles of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{added = }\text{0}\text{.0499}\text{L×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\\ {\text{Mole of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{ = 7}{\text{.485×10}}^{\text{-3}}\text{mol}{\text{hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$is calculated as follows,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{Initial}\text{amount}\text{base}\text{(mol)}\text{-}\text{Amount}\text{acid}\text{added}\text{(mol)}}{\text{Initial}\text{volume acid (L)}\text{+}\text{Volume base}\text{added}\text{(L)}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{0}\text{.00750}\text{mol}\text{-}\text{0}\text{.007485}\text{mol}}{\text{0}\text{.05}\text{L}\text{+}\text{0}\text{.0499}\text{L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{1}{\text{.5×10}}^{\text{-5}}\text{mol}}{\text{0}\text{.0999L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= 1}\text{.5}\times {\text{10}}^{-4}\text{M}\end{array}$

The $\text{pOH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pOH = -log}\left({\text{OH}}^{\text{-}}\right)\\ \text{pOH = -log}\left(1.5\times {\text{10}}^{-4}\right)\\ \text{pOH = 3}\text{.82}\end{array}$

The $\text{pH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pH + pOH =14}\text{.00}\\ \text{pH +3}\text{.82 =14}\text{.00}\\ \text{pH =10}\text{.17}\end{array}$

The $\text{pH}$ of the solution is calculated.

(d)

Interpretation Introduction

Interpretation:

After volume of $\text{50 mL}$ titrant has been added, the $\text{pH}$ of the solution has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

The titration of $\text{50 mL}$ of $\text{0}\text{.150 M HCl}$ solution with $\text{0}\text{.150 M NaOH}$ solution.

Volume of the acid solution is given below,

  $\begin{array}{l}\text{Total volume of the acid solution in liters =50 mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{Total volume of the solution in liters =}\text{0}\text{.05}\text{L}\end{array}$

Mole of hydronium ion is calculated as follows,

  $\begin{array}{l}{\text{Moles of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{added = }\text{0}\text{.05}\text{L×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\\ {\text{Mole of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{ = 7}{\text{.5×10}}^{\text{-3}}\text{mol}{\text{hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\end{array}$

The total moles of hydronium ion ${\text{ (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}$are equal to the concentration of hydroxide ions${\text{ (OH}}^{-}\text{)}$. Therefore the $\text{pH}$ of the solution is seven, and hence the solution is neutral.

(e)

Interpretation Introduction

Interpretation:

After volume of $\text{50}\text{.1 mL}$ titrant has been added, the $\text{pH}$ of the solution has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

The titration of $\text{50 mL}$ of $\text{0}\text{.150 M HCl}$ solution with $\text{0}\text{.150 M NaOH}$ solution.

Volume of the acid solution is given below,

  $\begin{array}{l}\text{Total volume of the acid solution in liters =50}\text{.1 mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{Total volume of the solution in liters =}\text{0}\text{.0501L}\end{array}$

Mole of hydronium ion is calculated as follows,

  $\begin{array}{l}{\text{Moles of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{added = }\text{0}\text{.0501}\text{L×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\\ {\text{Mole of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{ = 7}{\text{.515×10}}^{\text{-3}}\text{mol}{\text{hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$is calculated as follows,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{Initial}\text{amount}\text{base}\text{(mol)}\text{-}\text{Amount}\text{acid}\text{added}\text{(mol)}}{\text{Initial}\text{volume acid (L)}\text{+}\text{Volume base}\text{added}\text{(L)}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{0}\text{.007515}\text{mol}\text{-}\text{0}\text{.00750}\text{mol}}{\text{0}\text{.05}\text{L}\text{+}\text{0}\text{.0501}\text{L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{1}{\text{.5×10}}^{\text{-5}}\text{mol}}{\text{0}\text{.1001L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= 1}\text{.498}\times {\text{10}}^{-4}\text{M}\end{array}$

The $\text{pOH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pOH = -log}\left({\text{OH}}^{\text{-}}\right)\\ \text{pOH = -log}\left(1.498\times {\text{10}}^{-4}\right)\\ \text{pOH = 3}\text{.82}\end{array}$

The $\text{pH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pH + pOH =14}\text{.00}\\ \text{pH +3}\text{.82 =14}\text{.00}\\ \text{pH =10}\text{.17}\end{array}$

The $\text{pH}$ of the solution is calculated.

(f)

Interpretation Introduction

Interpretation:

After volume of $\text{75 mL}$ titrant has been added, the $\text{pH}$ of the solution has to be calculated.

Concept Introduction:

Refer to part (a)

Explanation of Solution

Given,

The titration of $\text{50 mL}$ of $\text{0}\text{.150 M HCl}$ solution with $\text{0}\text{.150 M NaOH}$ solution.

Volume of the acid solution is given below,

  $\begin{array}{l}\text{Total volume of the acid solution in liters =75 mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{Total volume of the solution in liters =}\text{0}\text{.075}\text{L}\end{array}$

Mole of hydronium ion is calculated as follows,

  $\begin{array}{l}{\text{Moles of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{added = }\text{0}\text{.075}\text{L×}\frac{\text{0}\text{.150}\text{mol}{\text{OH}}^{\text{-}}}{\text{1L}}\\ {\text{Mole of hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{ = 0}\text{.01125}\text{mol}{\text{hydronium ion (H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$is calculated as follows,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{Initial}\text{amount}\text{base}\text{(mol)}\text{-}\text{Amount}\text{acid}\text{added}\text{(mol)}}{\text{Initial}\text{volume acid (L)}\text{+}\text{Volume base}\text{added}\text{(L)}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{0}\text{.01125}\text{mol}\text{- 0}\text{.00750}\text{mol}}{\text{0}\text{.05}\text{L}\text{+}\text{0}\text{.075}\text{L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }\frac{\text{3}{\text{.75×10}}^{\text{-3}}\text{mol}}{\text{0}\text{.125L}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{= }0.03\text{M}\end{array}$

The $\text{pOH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pOH = -log}\left({\text{OH}}^{\text{-}}\right)\\ \text{pOH = -log}\left(0.03\right)\\ \text{pOH = 1}\text{.55}\end{array}$

The $\text{pH}$ of the solution is calculated as follows,

  $\begin{array}{l}\text{pH + pOH =14}\text{.00}\\ \text{pH +1}\text{.55=14}\text{.00}\\ \text{pH =12}\text{.45}\end{array}$

The $\text{pH}$ of the solution is calculated

(g)

Interpretation Introduction

Interpretation:

From the result titration curve has to be plotted and position of the equivalent point has to be indicated.

Explanation of Solution

Table for the volume of $\text{HCl}$ and $\text{pH}$ is given below,

S.No	Volume of $\text{HCl}$	$\text{pH}$
1	$\text{0}\text{.00 mL}$	$\text{13}\text{.17}$
2	$\text{0}\text{.25 mL}$	$\text{12}\text{.67}$
3	$\text{49}\text{.9 mL}$	$\text{10}\text{.17}$
4	$\text{50 mL}$	$7$
5	$\text{50}\text{.1 mL}$	$\text{10}\text{.17}$
6	$\text{75 mL}$	$\text{12}\text{.45}$

The equivalent point of the given solution is seven when the volume of $\text{HCl}$ is $\text{50}\text{.0}\text{mL}$.

From the result, the titration curve is plotted and position of the equivalent point is indicated as follows.