Chapter 9, Problem 9.10P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution formed from solution of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ and glycine amide $\left(\text{B}\right)$ has to be calculated.

Concept Introduction:

The equation for buffer is described by Henderson-Hasselbach equation and it is a rearranged form of equilibrium constant, $K_{\text{a}}$. It relates $\text{pH}$ of buffer solution and $\text{p}{K}_{\text{a}}$ value. It also helps to find equilibrium $\text{pH}$ in acid-base reactions. Consider an equation of dissociation of an acid as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Formula to calculate $\text{pH}$ is as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)$

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant of weak acid $\text{HA}$.

$\left[{\text{A}}^{-}\right]$ is concentration of conjugate base ${\text{A}}^{-}$

$\left[\text{HA}\right]$ is concentration of acid $\text{HA}$.

Concentration of a solution is expressed by molarity and is denoted by $M$. It is defined as the ratio of number of moles of solute to volume of solution in $1\text{ L}$.

The formula to calculate moles is as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

The formula used to calculate concentration is as follows:

  $\text{Concentration}=\frac{\text{moles}}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 9.10P

$\text{pH}$ of solution is 3.59.

Explanation of Solution

The formula to calculate concentration of glycine amide $\left(\text{B}\right)$ is as follows:

  $\left[\text{B}\right]=\frac{\left(\frac{\text{mass of B}}{\text{molar mass B}}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (1)

Substitute $1.00\text{ g}$ for mass of $\text{B}$, $74.08\text{ g/mol}$ for molar mass of $\text{B}$ and $0.100\text{ L}$ for volume of solution in equation (1).

  $\begin{array}{c}\left[\text{B}\right]=\frac{\left(\frac{1.00\text{ g}}{74.08\text{ g/mol}}\right)}{0.100\text{ L}}\\ =0.135\text{ M}\end{array}$

The formula to calculate concentration of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ is as follows:

  $\left[{\text{BH}}^{+}\right]=\frac{\left(\frac{{\text{mass of BH}}^{+}}{{\text{molar mass BH}}^{+}}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (2)

Substitute $1.00\text{ g}$ for mass of ${\text{BH}}^{+}$, $110.54\text{ g/mol}$ for molar mass of ${\text{BH}}^{+}$ and $0.100\text{ L}$ for volume of solution in equation (2).

  $\begin{array}{c}\left[{\text{BH}}^{+}\right]=\frac{\left(\frac{1.00\text{ g}}{110.54\text{ g/mol}}\right)}{0.100\text{ L}}\\ =0.090\text{ M}\end{array}$

Hence, concentration of glycine amide $\left(\text{B}\right)$ is $0.135\text{ M}$ and glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ is $0.090\text{ M}$. In accordance to Henderson-Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}\right)$        (3)

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant.

$\left[\text{B}\right]$ is concentration of $\text{B}$.

$\left[{\text{BH}}^{+}\right]$ is concentration of ${\text{BH}}^{+}$.

Substitute 8.20 for $\text{p}{K}_{\text{a}}$, $0.090\text{ M}$ for $\left[{\text{BH}}^{+}\right]$ and $0.135\text{ M}$ for $\left[\text{B}\right]$ in equation (3).

  $\begin{array}{c}\text{pH}=8.20+\log \left(\frac{0.135\text{ M}}{0.090\text{ M}}\right)\\ =8.37\end{array}$

Hence, $\text{pH}$ of solution is 8.37.

(b)

Interpretation Introduction

Interpretation:

Mass of glycine amide $\left(\text{B}\right)$ at $\text{pH}$ equals to 8 has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 9.10P

Mass of glycine amide $\left(\text{B}\right)$ is $0.423\text{ g}$.

Explanation of Solution

The formula to calculate concentration of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ is as follows:

  $\left[{\text{BH}}^{+}\right]=\frac{\left(\frac{{\text{mass of BH}}^{+}}{{\text{molar mass BH}}^{+}}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (2)

Substitute $1.00\text{ g}$ for mass of ${\text{BH}}^{+}$, $110.54\text{ g/mol}$ for molar mass of ${\text{BH}}^{+}$ and $100\text{ mL}$ for volume of solution in equation (2).

  $\begin{array}{c}\left[{\text{BH}}^{+}\right]=\frac{\left(\frac{1.00\text{ g}}{110.54\text{ g/mol}}\right)}{100\text{ mL}}\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\\ =0.090\text{ M}\end{array}$

In accordance to Henderson-Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}\right)$        (4)

Rearrange equation (4) for $\left[\text{B}\right]$.

  $\left[\text{B}\right]={10}^{\left(\text{pH}-\text{p}{K}_{\text{a}}+\log \left[{\text{BH}}^{+}\right]\right)}$        (5)

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant.

$\left[\text{B}\right]$ is concentration of $\text{B}$.

$\left[{\text{BH}}^{+}\right]$ is concentration of ${\text{BH}}^{+}$.

Substitute 8.20 for $\text{p}{K}_{\text{a}}$, $0.090\text{ M}$ for $\left[{\text{BH}}^{+}\right]$ and 8 for $\text{pH}$ in equation (3).

  $\begin{array}{c}\left[\text{B}\right]={10}^{\left(\text{8}-8.20+\log \left[0.090\text{ M}\right]\right)}\\ ={10}^{-1.25}\\ =0.057\text{ M}\end{array}$

The formula to calculate concentration of glycine amide $\left(\text{B}\right)$ is as follows:

  $\left[\text{B}\right]=\frac{\left(\frac{\text{mass of B}}{\text{molar mass B}}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (1)

Rearrange equation (1) for mass of $\text{B}$.

  $\text{Mass of B}=\left[\text{B}\right]\left(\text{volume of solution}\left(\text{L}\right)\right)\left(\text{molar mass of B}\right)$        (6)

Substitute $0.057\text{ M}$ for $\left[\text{B}\right]$, $74.08\text{ g/mol}$ for molar mass of $\text{B}$ and $100\text{ mL}$ for volume of solution in equation (6).

  $\begin{array}{c}\text{Mass of B}=\left(\text{0}\text{.0572 M}\right)\left(\text{74}\text{.08 g/mol}\right)\left(\text{100 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.423\text{ g}\end{array}$

Hence, mass of glycine amide $\left(\text{B}\right)$ is $0.423\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution formed from mixture of $\text{HCl}$ with solution of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ and glycine amide $\left(\text{B}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 9.10P

$\text{pH}$ of solution is 8.33.

Explanation of Solution

The formula to calculate number of moles of glycine amide $\left(\text{B}\right)$ is as follows:

  $\text{Moles}=\frac{\text{mass of B}}{\text{molar mass B}}$        (7)

Substitute $1.00\text{ g}$ for mass of $\text{B}$ and $74.08\text{ g/mol}$ for molar mass of in equation (7).

  $\begin{array}{c}\text{Moles of B}=\left(\frac{1.00\text{ g}}{74.08\text{ g/mol}}\right)\\ =0.0135\text{ mol}\end{array}$

The formula to calculate number of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ is as follows:

  ${\text{Moles of BH}}^{+}=\frac{{\text{mass of BH}}^{+}}{{\text{molar mass BH}}^{+}}$        (8)

Substitute $1.00\text{ g}$ for mass of ${\text{BH}}^{+}$ and $110.54\text{ g/mol}$ for molar mass of ${\text{BH}}^{+}$ in equation (8).

  $\begin{array}{c}{\text{Moles of BH}}^{+}=\frac{1.00\text{ g}}{110.54\text{ g/mol}}\\ =0.0090\text{ M}\end{array}$

The formula used to calculate concentration of $\text{HCl}$ is as follows:

  $\text{Concentration of HCl}=\frac{\text{moles of HCl}}{\text{volume of solution}\left(\text{L}\right)}$        (9)

Rearrange equation (9) for moles of $\text{HCl}$.

  $\text{Moles of HCl}=\left(\text{Concentration of HCl}\right)\left(\text{volume of solution}\left(\text{L}\right)\right)$        (10)

Substitute $0.100\text{ M}$ for concentration of $\text{HCl}$ and $5\text{ mL}$ for volume of solution in equation (10).

  $\begin{array}{c}\text{Moles of HCl}=\left(0.100\text{ M}\right)\left(5\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.0005\end{array}$

Solution formed of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ and glycine amide $\left(\text{B}\right)$ forms buffer and addition of $\text{HCl}$ produces conjugate acid. $\text{B}$ is denoted for base and ${\text{BH}}^{+}$ for its conjugate acid. Since volume is similar therefore, number of moles is equal to concentration. The ICE table for reaction is as follows:

$\begin{array}{cccccc}& \text{B}& +& {\text{H}}^{\text{+}}& \to & {\text{BH}}^{\text{+}}\\ \text{Initial concentration}\left(M\right)& 0.0135& & 0.0005& & 0.0090\\ \text{Change in concentration}\left(M\right)& -0.0005& & & & +0.0005\\ \text{final concentration}\left(M\right)& 0.013& & & & 0.0095\end{array}$

Hence, concentration of $\text{B}$ is $0.013\text{ M}$ and its conjugate acid ${\text{BH}}^{+}$ is $0.0095\text{ M}$. In accordance to Henderson-Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}\right)$        (11)

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant.

$\left[\text{B}\right]$ is concentration of $\text{B}$.

$\left[{\text{BH}}^{+}\right]$ is concentration of ${\text{BH}}^{+}$.

Substitute 8.20 for $\text{p}{K}_{\text{a}}$, $0.0095\text{ M}$ for $\left[{\text{BH}}^{+}\right]$ and $0.013\text{ M}$ for $\left[\text{B}\right]$ in equation (11).

  $\begin{array}{c}\text{pH}=8.20+\log \left(\frac{0.013\text{ M}}{0.0095\text{ M}}\right)\\ =8.33\end{array}$

Hence, $\text{pH}$ of solution is 8.33.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution formed from mixture of $\text{NaOH}$ with solution of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ and glycine amide $\left(\text{B}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 9.10P

$\text{pH}$ of solution is 8.41.

Explanation of Solution

The formula to calculate number of moles of glycine amide $\left(\text{B}\right)$ is as follows:

  $\text{Moles}=\frac{\text{mass of B}}{\text{molar mass B}}$        (7)

Substitute $1.00\text{ g}$ for mass of $\text{B}$ and $74.08\text{ g/mol}$ for molar mass of in equation (7).

  $\begin{array}{c}\text{Moles of B}=\left(\frac{1.00\text{ g}}{74.08\text{ g/mol}}\right)\\ =0.0135\text{ mol}\end{array}$

The formula to calculate number of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ is as follows:

  ${\text{Moles of BH}}^{+}=\frac{{\text{mass of BH}}^{+}}{{\text{molar mass BH}}^{+}}$        (8)

Substitute $1.00\text{ g}$ for mass of ${\text{BH}}^{+}$ and $110.54\text{ g/mol}$ for molar mass of ${\text{BH}}^{+}$ in equation (8).

  $\begin{array}{c}{\text{Moles of BH}}^{+}=\frac{1.00\text{ g}}{110.54\text{ g/mol}}\\ =0.0090\text{ M}\end{array}$

The formula used to calculate concentration of $\text{NaOH}$ is as follows:

  $\text{Concentration of NaOH}=\frac{\text{moles of NaOH}}{\text{volume of solution}\left(\text{L}\right)}$        (12)

Rearrange equation (12) for moles of $\text{NaOH}$.

  $\text{Moles of NaOH}=\left(\text{Concentration of NaOH}\right)\left(\text{volume of solution}\left(\text{L}\right)\right)$        (13)

Substitute $0.100\text{ M}$ for concentration of $\text{NaOH}$ and $10\text{ mL}$ for volume of solution in equation (13).

  $\begin{array}{c}\text{Moles of HCl}=\left(0.100\text{ M}\right)\left(10\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.001\end{array}$

Solution formed of glycine amide hydrochloride $\left({\text{BH}}^{+}\right)$ and glycine amide $\left(\text{B}\right)$ forms buffer and addition of $\text{NaOH}$ produces conjugate acid. Since volume is similar therefore, number of moles is equal to concentration. The ICE table for reaction is as follows:

$\begin{array}{cccccc}& {\text{BH}}^{\text{+}}& +& {\text{OH}}^{-}& \to & \text{B}\\ \text{Initial concentration}\left(M\right)& 0.0095& & 0.001& & 0.013\\ \text{Change in concentration}\left(M\right)& -0.001& & & & +0.0010\\ \text{final concentration}\left(M\right)& 0.00855& & & & 0.014\end{array}$

Hence, concentration of $\text{B}$ is $0.014\text{ M}$ and its conjugate acid ${\text{BH}}^{+}$ is $0.00855\text{ M}$. In accordance to Henderson-Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}\right)$        (14)

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant.

$\left[\text{B}\right]$ is concentration of $\text{B}$.

$\left[{\text{BH}}^{+}\right]$ is concentration of ${\text{BH}}^{+}$.

Substitute 8.20 for $\text{p}{K}_{\text{a}}$, $0.00855\text{ M}$ for $\left[{\text{BH}}^{+}\right]$ and $0.014\text{ M}$ for $\left[\text{B}\right]$ in equation (14).

  $\begin{array}{c}\text{pH}=8.20+\log \left(\frac{0.014\text{ M}}{0.00855\text{ M}}\right)\\ =8.41\end{array}$

Hence, $\text{pH}$ of solution is 8.41.