Chapter 16, Problem 16.23QE

Interpretation Introduction

Interpretation:

The pH for the titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{0, 0}\text{.50, 1}\text{.00, 2}\text{.40, and 3}{\text{.00 mL HNO}}_{\text{3}}$ has to be calculated and also the titration curve has to be sketched.

Explanation of Solution

Given,

    Titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{0, 0}\text{.50, 1}\text{.00, 2}\text{.40, and 3}{\text{.00 mL HNO}}_{\text{3}}$

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

Given system is titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of ${\text{0 mL HNO}}_{\text{3}}$

The concentration of $Ba{(OH)}_2$ is $2\times 0.24$= 0.48 M,

    $\begin{array}{l}\text{pOH = - log }\left({\text{OH}}^{-}\right)\\ \text{pOH = - log }\left(0.48\right)\\ \text{pOH = 0}\text{.32}\\ \\ \text{pH}\text{=}\text{pH +}\text{pOH}\text{=}\text{14}\\ \text{pH +}\text{0}\text{.32=}\text{14}\\ \text{pH =}\text{13}\text{.68}\end{array}$

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

Given system is titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{0}{\text{.50 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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 $Ba{(OH)}_2$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of Ba(OH)}}_{\text{2}}{\text{ = 1×10}}^{\text{-3}}\text{ L}{\text{of Ba(OH)}}_{\text{2}}\text{×}\frac{\text{0}\text{.24}\text{mol}{\text{Ba(OH)}}_{\text{2}}}{\text{1}\text{L}\text{of}{\text{Ba(OH)}}_{\text{2}}}\\ {\text{Moles of Ba(OH)}}_{\text{2}}\text{ = 2}{\text{.4×10}}^{\text{-4}}\text{moles}\end{array}$

Limiting reagent:

According to the balanced chemical equation, for 1 mole of $HN{O}_3$ needed 0.5 mol of $Ba{\left(OH\right)}_2$. So for $\text{1}\times {\text{10}}^{-4}\text{moles}$ of $HN{O}_3$, needed ($\text{1}\times {\text{10}}^{-4}\text{moles / 2}$) = $5\times {\text{10}}^{-5}\text{moles }$ of $Ba{\left(OH\right)}_2$.

But there is excess amount of $Ba{\left(OH\right)}_2$. Thus, $HN{O}_3$ is the limiting reagent. The limiting reagent will be fully used after the completion of the reaction.

ICE table:

  ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

S (mol)	$\text{1}\times {\text{10}}^{-4}\text{moles}$	$\text{2}{\text{.4×10}}^{\text{-4}}\text{moles}$	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) $\text{1}{\text{.5×10}}^{\text{-3}}\text{L}$

One mole of $Ba{\left(OH\right)}_2$ gives two moles of hydroxide ($O{H}^{-}$) ion, therefore,

$\text{1}\text{.4}\times {\text{10}}^{-4}\text{moles}$ of $Ba{\left(OH\right)}_2$ gives $\text{2×1}{\text{.4×10}}^{\text{-4}}\text{moles = 2}{\text{.8×10}}^{\text{-4}}\text{moles}$ of hydroxide ($O{H}^{-}$) ion.

  $\begin{array}{l}Concentrationof\text{ base }=\frac{Molesofexcessbase}{Totalvolume}\\ Concentrationof\text{ base }=\frac{\text{2}{\text{.8×10}}^{\text{-4}}\text{moles}}{\text{1}\text{.5}\times {\text{10}}^{-3}L}\\ Concentrationof\text{ base }=0.19\text{M}\end{array}$

The concentration of $Ba{\left(OH\right)}_2$ of 0.19 M,

    $\begin{array}{l}\text{pOH = - log }\left({\text{OH}}^{-}\right)\\ \text{pOH = - log }\left(0.19\right)\\ \text{pOH = 0}\text{.72}\\ \\ \text{pH}\text{=}\text{pH +}\text{pOH}\text{=}\text{14}\\ \text{pH +}0.72\text{=}\text{14}\\ \text{pH =}\text{13}\text{.28}\end{array}$

Addition of $\text{1}{\text{.0 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{1}{\text{.0 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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

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

Moles of $Ba{(OH)}_2$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of Ba(OH)}}_{\text{2}}{\text{ = 1×10}}^{\text{-3}}\text{ L}{\text{of Ba(OH)}}_{\text{2}}\text{×}\frac{\text{0}\text{.24}\text{mol}{\text{Ba(OH)}}_{\text{2}}}{\text{1}\text{L}\text{of}{\text{Ba(OH)}}_{\text{2}}}\\ {\text{Moles of Ba(OH)}}_{\text{2}}\text{ = 2}{\text{.4×10}}^{\text{-4}}\text{moles}\end{array}$

Limiting reagent:

According to the balanced chemical equation, for 1 mole of $HN{O}_3$ needed 0.5 mol of $Ba{\left(OH\right)}_2$. So for $\text{2}\times {\text{10}}^{-4}\text{moles}$ of $HN{O}_3$, needed ($\text{2}\times {\text{10}}^{-4}\text{moles / 2}$) = $1\times {\text{10}}^{-4}\text{moles }$ of $Ba{\left(OH\right)}_2$.

But there is excess amount of $Ba{\left(OH\right)}_2$. Thus, $HN{O}_3$ is the limiting reagent. The limiting reagent will be fully used after the completion of the reaction.

ICE table:

  ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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

The total volume of the given solution is 1 ml + 1 ml = 2 ml (or) ${\text{2×10}}^{\text{-3}}\text{L}$

One mole of $Ba{\left(OH\right)}_2$ gives two moles of hydroxide ($O{H}^{-}$) ion, therefore,

$\text{0}\text{.4}\times {\text{10}}^{-4}\text{moles}$ of $Ba{\left(OH\right)}_2$ gives $\text{2×0}{\text{.4×10}}^{\text{-4}}{\text{moles = 8×10}}^{\text{-5}}\text{moles}$ of hydroxide ($O{H}^{-}$) ion.

  $\begin{array}{l}Concentrationof\text{ base }=\frac{Molesofexcessbase}{Totalvolume}\\ Concentrationof\text{ base }=\frac{{\text{8×10}}^{\text{-5}}\text{moles}}{2\times {\text{10}}^{-3}L}\\ Concentrationof\text{ base }=0.04\text{M}\end{array}$

The concentration of $Ba{\left(OH\right)}_2$ of 0.04 M,

    $\begin{array}{l}\text{pOH = - log }\left({\text{OH}}^{-}\right)\\ \text{pOH = - log }\left(0.04\right)\\ \text{pOH = 1}\text{.40}\\ \\ \text{pH}\text{=}\text{pH +}\text{pOH}\text{=}\text{14}\\ \text{pH +}1.40\text{=}\text{14}\\ \text{pH =}\text{12}\text{.60}\end{array}$

Addition of $\text{2}{\text{.4 mL HNO}}_{\text{3}}$:

Given system is titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{2}{\text{.4 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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

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

Moles of $Ba{(OH)}_2$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of Ba(OH)}}_{\text{2}}{\text{ = 1×10}}^{\text{-3}}\text{ L}{\text{of Ba(OH)}}_{\text{2}}\text{×}\frac{\text{0}\text{.24}\text{mol}{\text{Ba(OH)}}_{\text{2}}}{\text{1}\text{L}\text{of}{\text{Ba(OH)}}_{\text{2}}}\\ {\text{Moles of Ba(OH)}}_{\text{2}}\text{ = 2}{\text{.4×10}}^{\text{-4}}\text{moles}\end{array}$

ICE table:

  ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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

The total volume of the given solution is 1 ml + 2.4 ml = 3.4 ml (or) $\text{3}{\text{.4×10}}^{\text{-3}}\text{L}$

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

Given system is titration of $\text{1}\text{.00 mL of 0}{\text{.240 M Ba(OH)}}_{\text{2}}\text{ with 0}{\text{.200 M HNO}}_{\text{3}}$ after addition of $\text{3}{\text{.0 mL HNO}}_{\text{3}}$.

The given titration is shown below,

    ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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

  $\begin{array}{l}{\text{Moles of HNO}}_{\text{3}}\text{ = 3}\times {\text{10}}^{-3}\text{ L}{\text{of HNO}}_{\text{3}}\text{×}\frac{\text{0}\text{.200}\text{mol}{\text{HNO}}_{\text{3}}}{\text{1}\text{L}\text{of}{\text{HNO}}_{\text{3}}}\\ {\text{Moles of HNO}}_{\text{3}}\text{ = 6}\times {\text{10}}^{-4}\text{moles}\end{array}$

Moles of $Ba{(OH)}_2$ is calculated as follows,

  $\begin{array}{l}{\text{Moles of Ba(OH)}}_{\text{2}}{\text{ = 1×10}}^{\text{-3}}\text{ L}{\text{of Ba(OH)}}_{\text{2}}\text{×}\frac{\text{0}\text{.24}\text{mol}{\text{Ba(OH)}}_{\text{2}}}{\text{1}\text{L}\text{of}{\text{Ba(OH)}}_{\text{2}}}\\ {\text{Moles of Ba(OH)}}_{\text{2}}\text{ = 2}{\text{.4×10}}^{\text{-4}}\text{moles}\end{array}$

Limiting reagent:

According to the balanced chemical equation, for 1 mole of $HN{O}_3$ needed 0.5 mol of $Ba{\left(OH\right)}_2$. So for $6\times {\text{10}}^{-4}\text{moles}$ of $HN{O}_3$, needed ($6\times {\text{10}}^{-4}\text{moles / 2}$) = $3\times {\text{10}}^{-4}\text{moles }$ of $Ba{\left(OH\right)}_2$.

But there is less amount of $Ba{\left(OH\right)}_2$. Thus, $Ba{\left(OH\right)}_2$ is the limiting reagent. The limiting reagent will be fully used after the completion of the reaction.

ICE table:

  ${\text{2HNO}}_{\text{3}}{\text{ (aq) + Ba(OH)}}_{\text{2}}\text{ (aq) }\to {\text{ Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ (aq) + 2H}}_{\text{2}}\text{O (l) }$

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

The total volume of the given solution is 1 ml + 3 ml = 4 ml (or) ${\text{4×10}}^{\text{-3}}\text{L}$

    $\begin{array}{l}Concentrationof\text{ acid }=\frac{Molesofexcessacid}{Totalvolume}\\ Concentrationof\text{ acid }=\frac{3.6\times {\text{10}}^{-4}M}{{\text{4×10}}^{\text{-3}}\text{L}}\\ Concentrationof\text{ acid }=\text{ 0}\text{.09M}\end{array}$

The concentration of $HN{O}_3$ of 0.09 M,

    $\begin{array}{l}\text{pH = - log }\left({\text{H}}^{+}\right)\\ \text{pH = - log }\left(0.09\right)\\ \text{pH = 1}\text{.05}\end{array}$

The titration curve is shown below,

Figure 1

The curve in Question 16.21 has the same general form, but the equivalence point occurs at half the volume the equivalence point occurs in Question 16.23.