Chapter 14, Problem 50QAP

Morphine, C₁₇H₁₉O₃N, is a weak base $\left(K_b=7.4\times {10}^{-7}\right)$. Consider its titration with hydrochloric acid. In the titration, 50.0 mL of a 0.1500 M solution of morphine is titrated with 0.1045 M HCl.

(a) Write a balanced net ionic equation for the reaction that takes place during titration.

(b) What are the species present at the equivalence point?

(c) What volume of hydrochloric acid is required to reach the equivalence point?

(d) What is the pH of the solution before any HCl is added?

(e) What is the pH of the solution halfway to the equivalence point?

(f) What is the pH of the solution at the equivalence point?

Interpretation Introduction

(a)

Interpretation:

The molarity of hydrochloric acid and morphine is 0.1045M and 0.1500M respectively. The titration of a solution of morphine which is a weak base of amount 50.00 mL is done with the hydrochloric acid. The value of the constant of equilibrium of base for morphine is $7.4\times {10}^{-7}$. The balanced net ionic equation of reaction during titration is to be written.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration.

Answer to Problem 50QAP

The balanced net ionic equation of reaction during titration is −

${\text{H}}^{+}(aq)+{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}(aq)\to {\text{HC}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{N}}^{+}(aq)$

Explanation of Solution

From the question −

Morphine C₁₇ H₁₉ O₃ N = base

Hydrochloric acid, HCl = acid which is represented as H+ ion

In this chemical reaction, one mole of acid is reacting with one mole of the base. The equation of this chemical reaction is represented as follows:

${\text{H}}^{+}(aq)+{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}(aq)\to {\text{HC}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{N}}^{+}(aq)$

Interpretation Introduction

(b)

Interpretation:

The species present at equivalence point are to be determined.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration.

Answer to Problem 50QAP

The species present at equivalence point - ${\text{HC}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{N}}^{+}(aq){\text{ and Cl}}^{-}(aq)$

Explanation of Solution

The actual equation of the chemical reaction can be represented as

$\text{HCl}(aq)+{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}(aq)\to {\text{HC}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{N}}^{+}(aq)+{\text{Cl}}^{-}(aq)$

At equivalence point

The consumption of all reactants occurs and products will remain only.

Therefore at the equivalence point, species are- ${\text{HC}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{N}}^{+}(aq){\text{ and Cl}}^{-}(aq)$

Interpretation Introduction

(c)

Interpretation:

The volume of hydrochloric acid for equivalence point is to be determined.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration.

The Molarity equation-

$\mathrm{Molarity}(M)=\frac{Numberofmolesofsolute(n)}{volume(V)}$

Answer to Problem 50QAP

The volume of hydrochloric acid is 71.77 mL which is required to reach at equivalence point.

Explanation of Solution

At the equivalence point, the reaction is −

$\text{HCl}(aq)+{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}(aq)\to {\text{HC}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{N}}^{+}(aq)+{\text{Cl}}^{-}(aq)$

In the chemical reaction, one mole of acid is reacting with one mole of the base. Therefore ration between acid and base is 1:1.

Number of moles of Morphine is calculated by using the given formula:

$\text{Molarity(M)=}\frac{\text{Number of moles of solute(n)}}{\text{volume(V)}}$ ……………… (1)

Given that-

Molarity of morphine = 0.1500M

Volume = 0.050L

Put the above values in Equ (1)

$\text{Number of moles of solute(n)=Molarity(M)×volume(V)}$

$\text{Number of moles of solute(n)}=\text{0}\text{.1500M×0}\text{.050L}$

Number of moles of morphine = 0.0075 mol

Now the volume of HCl is obtained by −

$\text{volume(V)=}\frac{\text{Number of moles of solute(n)}}{\text{Molarity(M)}}$ ……………….. (2)

Given that-

Molarity of HCl is = 0.1045M

Number of moles of HCl = Number of moles of Morphine = 0.0075mol

Put the above values in Equ (2)

$\text{volume(V)=}\frac{\text{0}\text{.0075mol}}{\text{0}\text{.1045M}}$

Volume of HCl = 0.07177 L or 71.77mL

Interpretation Introduction

(d)

Interpretation:

The pH value of the solution is to be determined before the addition of HCl.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration.

The equilibrium constant of acid-

$K_a=\frac{[A][B]}{[AB]}$

The pH value can be calculated as follows:

$\text{pH }={\text{ log}}_{\text{10}}\left[{\text{H}}^{+}\right]$

Similarly, the pOH value can be calculated as follows:

$\text{pOH }={\text{ log}}_{\text{10}}\left[{\text{OH}}^{-}\right]$

And,

$\text{pH +pOH }=14$

Answer to Problem 50QAP

The pH value of the solution is 10.52 before the addition of HCl.

Explanation of Solution

Before the addition of HCl −

The concentration of HCl or [OH^-] = concentration of base, morphine = 0.1500 M

The chemical reaction is represented as-

${\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}(aq)+{\text{H}}_{\text{2}}\text{O}\to {\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{+}(aq)+{\text{OH}}^{-}(aq)$

Therefore,

$K_b=\frac{{\text{[C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N]}}$ ……………… (1)

At equilibrium,

Molar concentration of $[{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{+}]=x$

Molar concentration of $[{\text{OH}}^{-}]=x$

Therefore,

Molar concentration of $[{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}]=0.1500-x$

Given that-

K_b = 7.4×10-7

Put the above values in Equ (1)

$7.4\times {10}^{-7}=\frac{x\times x}{0.1500-x}$

On solving −

The value of x = 3.33×10-4

So,

Molar concentration of $[{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{+}]=3.33\times {10}^{-4}$

Molar concentration of $[{\text{OH}}^{-}]=3.33\times {10}^{-4}$

Now, the pH value can be calculated as follows:

$\text{pOH}={\log }_{10}[{\text{OH}}^{-}]$ ……………. (2)

$[{\text{OH}}^{-}]=3.33\times {10}^{-4}$

$\text{pOH}={\log }_{10}[3.33\times {10}^{-4}]$

$\text{pOH}=3.48$

And

pH +pOH =14

Then ,

pH = 14 − 3.48 = 10.52

The pH of the solution = 10.52

Interpretation Introduction

(e)

Interpretation:

The pH value of the solution at halfway of the equivalence point is to be determined.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration.

The pH value can be calculated as follows:

$\text{pH }={\text{ log}}_{\text{10}}\left[{\text{H}}^{+}\right]$

Similarly, the pOH value can be calculated as follows:

$\text{pOH }={\text{ log}}_{\text{10}}\left[{\text{OH}}^{-}\right]$

And,

$\text{pH +pOH }=14$

Answer to Problem 50QAP

The pH value of the solution at halfway of the equivalence point is 7.87.

Explanation of Solution

At the halfway of the equivalence point is-

The concentration of OH or [OH^-] = The value of pK_b of the base

Now the pH of the solution is

$\mathrm{pOH}={\text{ log}}_{\text{10}}\left[{\text{OH}}^{-}\right]$ ……………. (1)

Given that −

K_b = 7.4×10-7

Put the value of [OH^-] in equation (1),

$\text{pOH }=lo{g}_{10}\left[7.4\times {10}^{-7}\right]$

$\text{pOH}=6.13$

And

pH +pOH =14

Then ,

pH = 14 −6.13= 7.87

The pH of the solution = 7.87

Interpretation Introduction

(f)

Interpretation:

The pH value of the solution at the equivalence point is to be determined.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration.

The molarity equation −

$\text{Molarity(M)=}\frac{\text{Number of moles of solute(n)}}{\text{volume(V)}}$

The equilibrium constant of base and acid can be related as-

$K_b=\frac{K_w}{K_a}$

The pH value can be calculated as follows:

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

Answer to Problem 50QAP

The pH value of the solution at the equivalence point is 4.54.

Explanation of Solution

The actual reaction is −

$\text{HCl}(aq)+{\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}\text{N}(aq)\to {\text{C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{+}(aq)+{\text{Cl}}^{-}(aq)$

In the chemical reaction, one mole of acid is reacting with one mole of the base. Therefore ration between acid and base is 1:1.

Number of moles of Morphine is calculated by using the given formula-

$\text{Molarity(M)=}\frac{\text{Number of moles of solute(n)}}{\text{volume(V)}}$ ……………… (1)

Given that-

Molarity of morphine = 0.1500M

Volume = 0.050L

Put the above values in Equ (1)

$\text{Number of moles of solute(n)=Molarity(M)×volume(V)}$

$\text{Number of moles of solute(n)=0}\text{.1500M×0}\text{.050L}$

Number of moles of morphine = 0.0075 mol

Now, the volume of HCl is obtained by −

$\text{volume(V)=}\frac{\text{Number of moles of solute(n)}}{\text{Molarity(M)}}$ ……………….. (2)

Given that-

Molarity of HCl is = 0.1045M

Number of moles of HCl = Number of moles of Morphine = 0.0075mol

Put the above values in equation (2)

$\text{volume(V)=}\frac{\text{0}\text{.0075mol}}{\text{0}\text{.1045M}}$

Volume of HCl = 0.07177 mL or 71.77mL

Now $\text{molarity of }\left[{\text{Cl}}^{\text{-}}\right]\text{=}\frac{\text{Number of moles of morphine}}{\text{total volume}}$

$\text{molarity of }\left[{\text{Cl}}^{\text{-}}\right]\text{=}\frac{\text{0}\text{.0075mol}}{\text{0}\text{.050L+0}\text{.07177L}}$

$\left[{\text{Cl}}^{\text{-}}\right]\text{=0}\text{.061M}$

The concentration of [H⁺] is calculated by −

${\text{[H}}^{\text{+}}\text{]=}\frac{{\text{[C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{\text{+}}\text{]}}{{\text{[Cl}}^{\text{-}}\text{]}}$

Also,

$K_a=\frac{K_w}{K_b}$

From above both equations −

$\frac{{\text{[C}}_{\text{17}}{\text{H}}_{\text{19}}{\text{O}}_{\text{3}}{\text{NH}}^{\text{+}}\text{]}}{{\text{[Cl}}^{\text{-}}\text{]}}=\frac{K_w}{K_b}$

Or,

$\frac{x^2}{0.061}=\frac{{10}^{-14}}{7.4\times {10}^{-7}}$

On calculation,

$\left[{\text{H}}^{+}\right]=2.87\times {10}^{-5}$

Now the pH of the solution −

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

$\text{pH}=-{\log }_{10}[2.87\times {10}^{-5}]$

Thus, the pH of the solution is 4.54.