Chapter 10, Problem 10.19P

Interpretation Introduction

Interpretation:

Equivalence volume and $\text{pH}$ at indicated volumes has to be calculated and plot of $\text{pH}$ against $V_{\text{a}}$ as to be made.

Concept Introduction:

Dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$  

Here,

$M_1$ denotes molarity of  $\text{HBr}$ solution.

$V_1$ denotes volume of $\text{HBr}$ solution.

$M_2$ denotes molarity of $\text{NaOH}$ solution.

$V_2$ denotes volume  of $\text{NaOH}$ solution.

Explanation of Solution

Dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$        (1)

Substitute  $\text{1}\text{.00 M}$ for $M_1$, $\text{0}\text{.1 M}$ for $M_2$ , $\text{100 mL}$ for $V_2$, in equation (1).

  $\left(1.00\text{ M}\right)V_1=\left(\text{0}\text{.1 M}\right)\left(\text{100 mL}\right)$        (2)

Rearrange equation (2) to calculate value of $V_1$.

  $\begin{array}{c}{V}_1=\frac{\left(\text{0}\text{.1 M}\right)\left(\text{100 mL}\right)}{\left(1.00\text{ M}\right)}\\ =10\text{ mL}\end{array}$

Thus, equivalence volume is $10\text{ mL}$.

Formula to calculate $\text{p}{K}_{\text{b}}$ from $\text{p}{K}_{\text{a}}$ is given as follows:

  $\text{p}{K}_{\text{b}}=14-\text{p}{K}_{\text{a}}$        (3)

Substitute 6.993 for $\text{p}{K}_{\text{b}}$ in equation (3).

  $\begin{array}{c}\text{p}{K}_{\text{b}}=14-6.993\\ =7.007\end{array}$

Formula to calculate $K_{\text{b}}$ from $\text{p}{K}_{\text{b}}$ is given as follows:

  $K_{\text{b}}={10}^{-\text{p}{K}_{\text{b}}}$        (4)

Substitute 7.007 for $\text{p}{K}_{\text{b}}$ in equation (4).

  $K_{\text{b}}={10}^{-7.007}$

Equilibrium for titration reaction is given as follows:

  ${\text{B}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons \text{HB}+{\text{OH}}^{-}$

Corresponding expression of $K_{\text{b}}$ is written as follows:

  $K_{\text{b}}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$        (4)

When no acid is added, ICE table is drawn as follows:

$\begin{array}{cccccc}& \text{B}& \rightleftharpoons & {\text{HB}}^{+}& +& {\text{OH}}^{-}\\ \text{Initial}& 0.050& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}& 0.050-x& & x& & x\end{array}$

Substitute $x$ to $\left[{\text{BH}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ , ${10}^{-7.007}$ for $K_{\text{b}}$ , $0.050-x$ for $\left[\text{B}\right]$ in equation (4).

  ${10}^{-7.007}=\frac{x^2}{0.050-x}$

Simplify to obtain the value of $x$ as $7.009\times {10}^{-5}\text{ M}$.

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

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (5)

Substitute $7.009\times {10}^{-5}\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ in equation (5).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(7.009\times {10}^{-5}\right)\\ =4.15\end{array}$

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

  $\text{pH}=14-\text{pOH}$        (6)

Substitute 4.15 for $\text{pOH}$ in equation (6).

  $\begin{array}{c}\text{pH}=\text{14}-4.15\\ =9.845\\ \approx 9.85\end{array}$

$\text{pH}$ is calculated by Henderson-HasselBalch equation given as follows:

  $\text{pH}=\text{p}{K}_{\text{b}}+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$        (7)

When volume added is $\text{1 mL}$,  initial and final moles are tabulated as follows:

$\begin{array}{cccccccc}& \text{B}& +& {\text{HNO}}_3& \rightleftharpoons & {\text{NO}}_3{}^{-}& +& {\text{BH}}^{+}\\ \text{Initial}& 0.00125& & 0.000125& & 0& & 0\\ \text{Final}& 0.001125& & 0& & 0.000125& & 0.000125\end{array}$

Substitute 7 for $\text{p}{K}_{\text{b}}$, $0.001125$ for $\left[\text{Salt}\right]$, $0.000125\text{ M}$ for $\left[\text{Acid}\right]$ in equation (7).

  $\begin{array}{c}\text{pH}=7+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ =7+\log \left(\frac{\text{0}\text{.009}}{\text{0}\text{.001}}\right)\\ =7.95\end{array}$

When volume added is $\text{5 mL}$,  initial and final moles are tabulated as follows:

$\begin{array}{cccccccc}& \text{B}& +& {\text{HNO}}_3& \rightleftharpoons & {\text{NO}}_3{}^{-}& +& {\text{BH}}^{+}\\ \text{Initial}& 0.00125& & 0.000625& & 0& & 0\\ \text{Final}& 0.000625& & 0& & 0.000625& & 0.000625\end{array}$

Substitute 7 for $\text{p}{K}_{\text{b}}$, $\text{0}\text{.000625 M}$ for $\left[\text{Salt}\right]$, $\text{0}\text{.000625 M}$ for $\left[\text{Acid}\right]$ in equation (7).

  $\begin{array}{c}\text{pH}=7+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ =7+\log \left(\frac{\text{0}\text{.000625}}{\text{0}\text{.000625}}\right)\\ =7\end{array}$

When volume added is $\text{9 mL}$,  initial and final moles are tabulated as follows:

$\begin{array}{cccccccc}& \text{B}& +& {\text{HNO}}_3& \rightleftharpoons & {\text{NO}}_3{}^{-}& +& {\text{BH}}^{+}\\ \text{Initial}& 0.00125& & 0.001125& & 0& & 0\\ \text{Final}& 0.000125& & 0& & 0.001125& & 0.001125\end{array}$

Substitute 7 for $\text{p}{K}_{\text{b}}$, $\text{0}\text{.000125 M}$ for $\left[\text{Salt}\right]$, $\text{0}\text{.001125 M}$ for $\left[\text{Acid}\right]$ in equation (7).

  $\begin{array}{c}\text{pH}=7+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ =7+\log \left(\frac{\text{0}\text{.000125}}{\text{0}\text{.001125}}\right)\\ =6.04\end{array}$

When volume added is $\text{9}\text{.90 mL}$,  initial and final moles are tabulated as follows:

$\begin{array}{cccccccc}& \text{B}& +& {\text{HNO}}_3& \rightleftharpoons & {\text{NO}}_3{}^{-}& +& {\text{BH}}^{+}\\ \text{Initial}& 0.00125& & 0.0012375& & 0& & 0\\ \text{Final}& 0.0000125& & 0& & 0.0012375& & 0.0012375\end{array}$

Substitute 7 for $\text{p}{K}_{\text{b}}$, $\text{0}\text{.0000125 M}$ for $\left[\text{Salt}\right]$, $\text{0}\text{.0012375 M}$ for $\left[\text{Acid}\right]$ in equation (7).

  $\begin{array}{c}\text{pH}=7+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ =7+\log \left(\frac{\text{0}\text{.0000125}}{\text{0}\text{.0012375}}\right)\\ =5.00\end{array}$

When volume added is $\text{10}\text{.00 mL}$,  initial and final moles are tabulated as follows:

$\begin{array}{cccccccc}& \text{B}& +& {\text{HNO}}_3& \rightleftharpoons & {\text{NO}}_3{}^{-}& +& {\text{BH}}^{+}\\ \text{Initial}& 0.00125& & 0.00125& & 0& & 0\\ \text{Final}& 0& & 0& & 0.00125& & 0.00125\end{array}$

Thus excess $\left[{\text{BH}}^{+}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{BH}}^{+}\right]=\frac{0.00125}{0.025+0.01}\\ =0.0357\text{ M}\end{array}$

Corresponding ICE table is as follows:

$\begin{array}{cccccc}& {\text{BH}}^{+}& \rightleftharpoons & \text{B}& +& {\text{H}}_3{\text{O}}^{+}\\ \text{Initial}& 0.0357& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}& 0.0357-x& & x& & x\end{array}$

Corresponding expression of $K_{\text{a}}$ is written as follows:

  $K_{\text{a}}=\frac{\left[\text{B}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[{\text{BH}}^{+}\right]}$        (8)

Substitute $x$ to $\left[\text{B}\right]$ and $\left[{\text{H}}_3{\text{O}}^{+}\right]$ , $1.02\times {10}^{-7}$ for $K_{\text{a}}$ , $0.0357-x$ for $\left[{\text{BH}}^{+}\right]$ in equation (4).

  $1.02\times {10}^{-7}=\frac{x^2}{0.0357-x}$

Simplify to obtain the value of $x$ as $6.029\times {10}^{-5}$.

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

  $\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$        (9)

Substitute $6.029\times {10}^{-5}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (9).

  $\begin{array}{c}\text{pH}=-\text{log}\left(6.029\times {10}^{-5}\right)\\ =4.219\\ \approx 4.22\end{array}$

When volume added is $\text{10}\text{.10 mL}$, excess $\left[{\text{H}}^{+}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{\left(0.001265-0.00125\right)\text{ mol}}{\left(0.025+0.0101\right)\text{ L}}\\ =3.56\times {10}^{-4}\text{ M}\end{array}$

Substitute $3.56\times {10}^{-4}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (9).

  $\begin{array}{c}\text{pH}=-\text{log}\left(3.56\times {10}^{-4}\right)\\ =3.448\\ \approx 3.45\end{array}$

When volume added is $\text{12}\text{.00 mL}$, excess $\left[{\text{H}}^{+}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{\left(0.0015-0.00125\right)\text{ mol}}{\left(0.025+0.012\right)\text{ L}}\\ =0.00675\text{ M}\end{array}$

Substitute $0.00675\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (9).

  $\begin{array}{c}\text{pH}=-\text{log}\left(0.00675\right)\\ =2.17\end{array}$

Corresponding plot of $\text{pH}$ against $V_{\text{a}}$ is drawn as follows: