Chapter 16, Problem 16.21QE

Interpretation Introduction

Interpretation:

The pH of the titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition $\text{0, 0}\text{.25, 0}\text{.50, 1}\text{.20, and 1}{\text{.50 mL HNO}}_{\text{3}}$ has to be calculated and also the titration curve has to be sketched.

Explanation of Solution

Given,

    Titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{0, 0}\text{.25, 0}\text{.50, 1}\text{.20, and 1}{\text{.50 mL HNO}}_{\text{3}}$

Addition of ${\text{0 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of ${\text{0 mL HNO}}_{\text{3}}$

The concentration of $LiOH$ 0.24 M,

    $\begin{array}{l}\text{pOH = - log }\left({\text{OH}}^{-}\right)\\ \text{pOH = - log }\left(0.24\right)\\ \text{pOH = 0}\text{.62}\\ \\ \text{pH}\text{=}\text{pH +}\text{pOH}\text{=}\text{14}\\ \text{pH +}\text{0}\text{.62=}\text{14}\\ \text{pH =}\text{13}\text{.38}\end{array}$

Addition of $\text{0}{\text{.25 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{0}{\text{.25 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}{\text{ (aq) + H}}_{\text{2}}\text{O (l) }$

Moles of $HN{O}_3$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of HNO}}_{\text{3}}\text{ = 2}\text{.5}\times {\text{10}}^{-4}\text{ L}{\text{of HNO}}_{\text{3}}\text{×}\frac{\text{0}\text{.200}\text{mol}{\text{HNO}}_{\text{3}}}{\text{1}\text{L}\text{of}{\text{HNO}}_{\text{3}}}\\ {\text{Moles of HNO}}_{\text{3}}\text{ = 5}\times {\text{10}}^{-5}\text{moles}\end{array}$

Moles of $\text{LiOH}$ is calculated as follows,

  $\begin{array}{l}\text{Moles of LiOH = 1}\times {\text{10}}^{-3}\text{ L}\text{of LiOH}\text{×}\frac{\text{0}\text{.24}\text{mol}\text{LiOH}}{\text{1}\text{L}\text{of}\text{LiOH}}\\ \text{Moles of LiOH}\text{ = 2}\text{.4}\times {\text{10}}^{-4}\text{moles}\end{array}$

ICE table:

  ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}\text{ (aq) }\text{ + }{\text{H}}_{\text{2}}\text{O (l) }$

S (mol)	$\text{5}\times {\text{10}}^{-5}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	0	Excess
R (mol)	$\text{-5}\times {\text{10}}^{-5}\text{moles}$	$\text{-5}\times {\text{10}}^{-5}\text{moles}$	$\text{5}\times {\text{10}}^{-5}\text{moles}$	$\text{5}\times {\text{10}}^{-5}\text{moles}$
F(mol)	0	$\text{1}\text{.9}\times {\text{10}}^{-4}\text{moles}$	$\text{5}\times {\text{10}}^{-5}\text{moles}$	Excess

The total volume of the given solution is 1 ml + 0.25 ml = 1.25 ml (or) 0.0625 L

  $\begin{array}{l}Concentrationof\text{ base }=\frac{Molesofexcessbase}{Totalvolume}\\ Concentrationof\text{ base }=\frac{1.9\times {10}^{-4}M}{0.00125L}\\ Concentrationof\text{ base }=\text{ 0}\text{.152M}\end{array}$

The concentration of $LiOH$ 0.152 M,

    $\begin{array}{l}\text{pOH = - log }\left({\text{OH}}^{-}\right)\\ \text{pOH = - log }\left(0.152\right)\\ \text{pOH = 0}\text{.818}\\ \\ \text{pH}\text{=}\text{pH +}\text{pOH}\text{=}\text{14}\\ \text{pH +}0.818\text{=}\text{14}\\ \text{pH =}\text{13}\text{.18}\end{array}$

Addition of $\text{0}{\text{.50 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{0}{\text{.50 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}{\text{ (aq) + H}}_{\text{2}}\text{O (l) }$

Moles of $HN{O}_3$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of HNO}}_{\text{3}}\text{ = 5}\times {\text{10}}^{-4}\text{ L}{\text{of HNO}}_{\text{3}}\text{×}\frac{\text{0}\text{.200}\text{mol}{\text{HNO}}_{\text{3}}}{\text{1}\text{L}\text{of}{\text{HNO}}_{\text{3}}}\\ {\text{Moles of HNO}}_{\text{3}}\text{ = 1}\times {\text{10}}^{-4}\text{moles}\end{array}$

Moles of $\text{LiOH}$ is calculated as follows,

  $\begin{array}{l}\text{Moles of LiOH = 1}\times {\text{10}}^{-3}\text{ L}\text{of LiOH}\text{×}\frac{\text{0}\text{.24}\text{mol}\text{LiOH}}{\text{1}\text{L}\text{of}\text{LiOH}}\\ \text{Moles of LiOH}\text{ = 2}\text{.4}\times {\text{10}}^{-4}\text{moles}\end{array}$

ICE table:

  ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}\text{ (aq) }\text{ + }{\text{H}}_{\text{2}}\text{O (l) }$

S (mol)	$\text{1}\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	0	Excess
R (mol)	$\text{-1}\times {\text{10}}^{-4}\text{moles}$	$\text{-1}\times {\text{10}}^{-4}\text{moles}$	$\text{1}\times {\text{10}}^{-4}\text{moles}$	$\text{1}\times {\text{10}}^{-4}\text{moles}$
F(mol)	0	$\text{1}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{1}\times {\text{10}}^{-4}\text{moles}$	Excess

The total volume of the given solution is 1 ml + 0.50 ml = 1.50 ml (or) 0.0015 L

  $\begin{array}{l}Concentrationof\text{ base }=\frac{Molesofexcessbase}{Totalvolume}\\ Concentrationof\text{ base }=\frac{1.4\times {10}^{-4}M}{0.0015L}\\ Concentrationof\text{ base }=\text{ 0}\text{.0933M}\end{array}$

The concentration of $LiOH$ 0.152 M,

    $\begin{array}{l}\text{pOH = - log }\left({\text{OH}}^{-}\right)\\ \text{pOH = - log }\left(0.0933\right)\\ \text{pOH = 1}\text{.03}\\ \\ \text{pH}\text{=}\text{pH +}\text{pOH}\text{=}\text{14}\\ \text{pH +}1.030\text{=}\text{14}\\ \text{pH =}\text{12}\text{.97}\end{array}$

Addition of $\text{1}{\text{.20 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{1}{\text{.20 mL HNO}}_{\text{3}}$

The given titration is shown below,

    ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}{\text{ (aq) + H}}_{\text{2}}\text{O (l) }$

Moles of $HN{O}_3$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of HNO}}_{\text{3}}\text{ = 1}\text{.2}\times {\text{10}}^{-3}\text{ L}{\text{of HNO}}_{\text{3}}\text{×}\frac{\text{0}\text{.200}\text{mol}{\text{HNO}}_{\text{3}}}{\text{1}\text{L}\text{of}{\text{HNO}}_{\text{3}}}\\ {\text{Moles of HNO}}_{\text{3}}\text{ = 2}\text{.4}\times {\text{10}}^{-4}\text{moles}\end{array}$

Moles of $\text{LiOH}$ is calculated as follows,

  $\begin{array}{l}\text{Moles of LiOH = 1}\times {\text{10}}^{-3}\text{ L}\text{of LiOH}\text{×}\frac{\text{0}\text{.24}\text{mol}\text{LiOH}}{\text{1}\text{L}\text{of}\text{LiOH}}\\ \text{Moles of LiOH}\text{ = 2}\text{.4}\times {\text{10}}^{-4}\text{moles}\end{array}$

ICE table:

  ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}\text{ (aq) }\text{ + }{\text{H}}_{\text{2}}\text{O (l) }$

S (mol)	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	0	Excess
R (mol)	$\text{-2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{-2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$
F(mol)	0	0	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	Excess

The total volume of the given solution is 1 ml + 1.2 ml = 2.20 ml (or) 0.0022 L

The resulting solution is neutral. Therefore,

    ${\text{K}}_{\text{w}}\text{= 1}\text{.0}\times {\text{10}}^{\text{-14}}$

Hence,

    ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [OH}}^{-}\text{] = 1}\text{.0}\times {\text{10}}^{\text{-7}}\text{ M}$

The concentration of $HCl$ 0.08 M,

    $\begin{array}{l}\text{pH = - log }\left({\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right)\\ \text{pH = 7}\end{array}$

Addition of $\text{1}{\text{.5 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}\text{.240 M LiOH with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{1}{\text{.5 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}{\text{ (aq) + H}}_{\text{2}}\text{O (l) }$

Moles of $HN{O}_3$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of HNO}}_{\text{3}}\text{ = 1}\text{.5}\times {\text{10}}^{-3}\text{ L}{\text{of HNO}}_{\text{3}}\text{×}\frac{\text{0}\text{.200}\text{mol}{\text{HNO}}_{\text{3}}}{\text{1}\text{L}\text{of}{\text{HNO}}_{\text{3}}}\\ {\text{Moles of HNO}}_{\text{3}}\text{ = 3}\times {\text{10}}^{-4}\text{moles}\end{array}$

Moles of $\text{LiOH}$ is calculated as follows,

  $\begin{array}{l}\text{Moles of LiOH = 1}\times {\text{10}}^{-3}\text{ L}\text{of LiOH}\text{×}\frac{\text{0}\text{.24}\text{mol}\text{LiOH}}{\text{1}\text{L}\text{of}\text{LiOH}}\\ \text{Moles of LiOH}\text{ = 2}\text{.4}\times {\text{10}}^{-4}\text{moles}\end{array}$

ICE table:

  ${\text{HNO}}_{\text{3}}\text{ (aq) + LiOH (aq) }\to {\text{ LiNO}}_{\text{3}}\text{ (aq) }\text{ + }{\text{H}}_{\text{2}}\text{O (l) }$

S (mol)	$3\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	0	Excess
R (mol)	$\text{-2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{-2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$
F(mol)	$6\times {\text{10}}^{-5}\text{moles}$	0	$\text{2}\text{.4}\times {\text{10}}^{-4}\text{moles}$	Excess

The total volume of the given solution is 1 ml + 1.5 ml = 2.5 ml (or) 0.0025 L

  $\begin{array}{l}Concentrationof\text{ acid }=\frac{Molesofexcessacid}{Totalvolume}\\ Concentrationof\text{ acid }=\frac{6\times {\text{10}}^{-5}M}{0.0025L}\\ Concentrationof\text{ acid }=\text{ 0}\text{.024M}\end{array}$

The concentration of $HN{O}_3$ 0.152 M,

    $\begin{array}{l}\text{pH = - log }\left({\text{H}}^{+}\right)\\ \text{pH = - log }\left(0.024\right)\\ \text{pH = 1}\text{.62}\end{array}$

The titration curve is given below,

Figure 1