Chapter 6, Problem 6H.3E

(a)

Interpretation Introduction

Interpretation:

The required volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ to neutralize one-half of 25.0 mL of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$ has to be calculated.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

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

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

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

  ${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

Answer to Problem 6H.3E

The required volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ to neutralize one-half of 25.0 mL of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$ is $9.17\times {\text{10}}^{\text{-3}}\text{L}$.

Explanation of Solution

The amount of ${\text{OH}}^{-}$ ions in original analyte solution is,

$\begin{array}{l}{\text{n(OH}}^{\text{-}}\text{)}\text{=}\text{volume}\text{×}{\text{(OH}}^{\text{-}}\text{concentration)}\\ \\ \text{=}\text{(}\frac{\text{25}}{\text{2}}{\text{×10}}^{\text{-3}}\text{L)×(0}\text{.110}\text{mol}\cdot {\text{L}}^{\text{-1}}\text{)}\\ \\ \text{=}1.375\times {\text{10}}^{\text{-3}}\text{mol}\end{array}$

The chemical reaction for the neutralization process is given below.

  ${\text{HCl}}_{\text{(aq)}}{\text{+NaOH}}_{\text{(aq)}}\to {\text{NaCl}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

From the above reaction stoichiometry, $\text{1}\text{mol}\text{NaOH}\text{=}\text{1}\text{mol}\text{HCl}$, the same as $\text{1}\text{mol}{\text{OH}}^{-}\text{=}\text{1}\text{mol}{\text{H}}_3{\text{O}}^{+}$ from this information we can calculate the volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ required to neutralize one-half of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$.

Here, $\text{1}\text{.375}\times {\text{10}}^{\text{-3}}\text{mol}$${\text{OH}}^{-}$ = $\text{1}\text{.375}\times {\text{10}}^{\text{-3}}\text{mol}$${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

$\begin{array}{l}{\text{n(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{=}\text{volume}\text{×}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{concentration)}\\ \\ 1.375\times {\text{10}}^{\text{-3}}\text{mol}\text{=}\text{x×(0}\text{.150}\text{mol}\cdot {\text{L}}^{\text{-1}})\\ \\ \text{x}\text{=}\frac{1.375\times {\text{10}}^{\text{-3}}\text{mol}}{\text{0}\text{.150}\text{mol}\cdot {\text{L}}^{\text{-1}}}\\ \\ =9.17\times {\text{10}}^{\text{-3}}\text{L}\end{array}$

Hence, the required volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ to neutralize one-half of 25.0 mL of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$ is $9.17\times {\text{10}}^{\text{-3}}\text{L}$.

(b)

Interpretation Introduction

Interpretation:

The required volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ to neutralize all the hydroxide ions in 25.0 mL of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$ has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6H.3E

The required volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ to neutralize all the hydroxide ions in 25.0 mL of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$ is $18.33\times {\text{10}}^{\text{-3}}\text{L}$.

Explanation of Solution

The amount of ${\text{OH}}^{-}$ ions in original analyte solution is,

$\begin{array}{l}{\text{n(OH}}^{\text{-}}\text{)}\text{=}\text{volume}\text{×}{\text{(OH}}^{\text{-}}\text{concentration)}\\ \\ \text{=}{\text{(25×10}}^{\text{-3}}\text{L)×(0}\text{.110}\text{mol}\cdot {\text{L}}^{\text{-1}}\text{)}\\ \\ \text{=}\text{2}\text{.75}\times {\text{10}}^{\text{-3}}\text{mol}\end{array}$

The chemical reaction for the neutralization process is given below

  ${\text{HCl}}_{\text{(aq)}}{\text{+NaOH}}_{\text{(aq)}}\to {\text{NaCl}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

From the above reaction stoichiometry, $\text{1}\text{mol}\text{NaOH}\text{=}\text{1}\text{mol}\text{HCl}$, the same as $\text{1}\text{mol}{\text{OH}}^{-}\text{=}\text{1}\text{mol}{\text{H}}_3{\text{O}}^{+}$ from this information we can calculate the volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ required to neutralize one-half of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$.

Here, $\text{2}\text{.75}\times {\text{10}}^{\text{-3}}\text{mol}$${\text{OH}}^{-}$ = $\text{2}\text{.75}\times {\text{10}}^{\text{-3}}\text{mol}$${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

  $\begin{array}{l}{\text{n(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{=}\text{volume}\text{of}\text{HCl}\text{×}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{concentration)}\\ \\ 2.75\times {\text{10}}^{\text{-3}}\text{mol}\text{=}\text{x×(0}\text{.150}\text{mol}\cdot {\text{L}}^{\text{-1}})\\ \\ \text{x}\text{=}\frac{\text{2}\text{.75}\times {\text{10}}^{\text{-3}}\text{mol}}{\text{0}\text{.150}\text{mol}\cdot {\text{L}}^{\text{-1}}}\\ \\ =18.33\times {\text{10}}^{\text{-3}}\text{L}\end{array}$

Therefore, the required volume of $\text{0}\text{.150}\text{M}{\text{HCl}}_{\text{(aq)}}$ to neutralize all the hydroxide ions in 25.0 mL of $\text{0}\text{.110}\text{M}{\text{NaOH}}_{\text{(aq)}}$ is $18.33\times {\text{10}}^{\text{-3}}\text{L}$.

(c)

Interpretation Introduction

Interpretation:

The molar concentration of ${\text{Na}}^{\text{+}}$ ions at the stoichiometric point has to calculated.

Concept introduction:

Refer to part (a).

Explanation of Solution

The molar concentration of ${\text{Na}}^{\text{+}}$ ions at the stoichiometric point is

The amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions are calculated using the volume of titrant.

Existing volume of $\text{HCl}$ is 5.0 mL

Further addition of $\text{HCl}$ is 5.0 mL

Therefore, the total amount of $\text{HCl}$ is $\text{10}\text{.0 mL =}\text{10}{\text{.0×10}}^{\text{-3}}\text{mmol}$

  $\begin{array}{l}{\text{n(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{=}\text{(10}{\text{.0×10}}^{\text{-3}}\text{L)×(0}\text{.340}\text{mol}\cdot {\text{L}}^{\text{-1}}\text{)}\\ \\ \text{=}\text{3}{\text{.4×10}}^{\text{-3}}\text{mol}\\ \\ \text{=}\text{3}\text{.4}\text{mmol}\end{array}$

Hence, the amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the titrant is 3.4 mmol.

The unreacted ${\text{OH}}^{-}$ ions present in the analyte solution is

  $\begin{array}{l}{\text{n(OH}}^{\text{-}}{\text{)}}_{\text{final}}\text{=}{\text{V}}_{\text{initial}}{\text{-V}}_{\text{added}}\\ \\ \text{=}\text{6}\text{.25-3}\text{.4}\text{mmol }\\ \\ \text{=}\text{2}\text{.84}\text{mmol}\end{array}$

The molar concentration of ${\text{OH}}^{-}$ ion present in the total volume of solution is

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{\text{unreacted}{\text{OH}}^{\text{-}}\text{present}\text{in}\text{the}\text{analyte}\text{solution}}{{\text{V}}_{\text{analyte}}{\text{+V}}_{\text{titrant}}}\\ \\ \text{=}\frac{\text{2}{\text{.84×10}}^{\text{-3}}\text{mol}}{\text{(30}\text{.0+5}{\text{.0)×10}}^{\text{-3}}\text{L}}\\ \\ \text{=}\text{0}\text{.081}\text{mol}\cdot {\text{L}}^{\text{-1}}\end{array}$

Therefore, the concentration of ${\text{[OH}}^{-}]$ is $\text{0}\text{.081}\text{mol}\cdot {\text{L}}^{\text{-1}}$

Here, base is in excess, therefore, the $\text{pOH}$ of the solution is calculated, from that the pH of the solution is determined.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{] }\\ \\ \text{=}\text{-log(0}\text{.081)}\\ \\ \text{=}\text{1}\text{.09}\\ \\ \text{pH}\text{=}\text{14-pOH}\\ \\ \text{=}\text{14-1}\text{.09}\\ \\ \text{=}\text{12}\text{.91}\end{array}$

Hence, the $\text{pH}$ of the solution is 12.91.

The amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the volume of titrant is calculated

The

The amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions are calculated using the volume of titrant.

Existing volume of $\text{HCl}$ is 5.0 mL

Further addition of $\text{HCl}$ is 5.0 mL

Therefore, the total amount of $\text{HCl}$ is $\text{10}\text{.0 mL =}\text{10}{\text{.0×10}}^{\text{-3}}\text{mmol}$

  $\begin{array}{l}{\text{n(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{=}\text{(10}{\text{.0×10}}^{\text{-3}}\text{L)×(0}\text{.340}\text{mol}\cdot {\text{L}}^{\text{-1}}\text{)}\\ \\ \text{=}\text{3}{\text{.4×10}}^{\text{-3}}\text{mol}\\ \\ \text{=}\text{3}\text{.4}\text{mmol}\end{array}$

Hence, the amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the titrant is 3.4 mmol.

The unreacted ${\text{OH}}^{-}$ ions present in the analyte solution is

  $\begin{array}{l}{\text{n(OH}}^{\text{-}}{\text{)}}_{\text{final}}\text{=}{\text{V}}_{\text{initial}}{\text{-V}}_{\text{added}}\\ \\ \text{=}\text{6}\text{.25-3}\text{.4}\text{mmol }\\ \\ \text{=}\text{2}\text{.84}\text{mmol}\end{array}$