Chapter 13, Problem 13.17P

(a)

Interpretation Introduction

Interpretation:

From mass balance 13.15, the equation 13-16 has to be derived.

Concept Introduction:

Ion-pairing in acid-base systems:

The possible ion-pair equilibria in the mixture of sodium hydrogen tartrate $({\text{Na}}^{+}{\text{HT}}^{-})$, pyridinium chloride $({\text{PyH}}^{\text{+}}{\text{Cl}}^{\text{-}}\text{)}$ and $\text{KOH}$ are:

$\begin{array}{l}{\text{Na}}^{\text{+}}{\text{ + T}}^{{\text{2}}^{\text{-}}}\stackrel{}{\rightleftharpoons }{\text{ NaT}}^{\text{-}}{\text{K}}_{\text{NaT}}=\frac{{\text{[NaT}}^{\text{-}}]}{[{\text{Na}}^{\text{+}}{\text{][T}}^{\text{2-}}\text{]}}=8\mathrm{......13}-12\\ {\text{Na}}^{\text{+}}{\text{ + HT}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{ NaHT}{\text{K}}_{\text{NaT}}=\frac{\text{[NaHT}]}{[{\text{Na}}^{\text{+}}{\text{][HT}}^{\text{-}}\text{]}}=1.6\mathrm{......}\text{13-13}\end{array}$

Explanation of Solution

Derivation of equation 13-16 from mass balance 13-15:

Equation 13-15 (mass balance for sodium) is:

${\text{[Na}}^{\text{+}}{\text{] + [NaT}}^{\text{-}}{\text{] + [NaHT] = F}}_{\text{Na}}{\text{ = F}}_{{\text{H}}_{\text{2}}\text{T}}\text{ = 0}\text{.0200 M}\mathrm{.......}\text{ 13-15}$

Equation 13-16 is:

$[\text{Na+] = }\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{1+{\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[T}}^{\text{2-}}{\text{] + K}}_{\text{NaHT}}{\text{[HT}}^{\text{-}}\text{]}}$

Consider the two equations,

$\begin{array}{l}{\text{Na}}^{\text{+}}{\text{ + T}}^{{\text{2}}^{\text{-}}}\stackrel{}{\rightleftharpoons }{\text{ NaT}}^{\text{-}}{\text{K}}_{\text{NaT}}=\frac{{\text{[NaT}}^{\text{-}}]}{[{\text{Na}}^{\text{+}}{\text{][T}}^{\text{2-}}\text{]}}=8\mathrm{......13}-12\\ {\text{Na}}^{\text{+}}{\text{ + HT}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{ NaHT}{\text{K}}_{\text{NaT}}=\frac{\text{[NaHT}]}{[{\text{Na}}^{\text{+}}{\text{][HT}}^{\text{-}}\text{]}}=1.6\mathrm{......}\text{13-13}\end{array}$

Substitute ${\text{[NaT}}^{-}]={\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{][T}}^{\text{2-}}{\text{] and [NaHT] = K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}{\text{][HT}}^{\text{-}}\text{]}$

$\begin{array}{l}{\text{[Na+]+K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{][T}}^{\text{2-}}\text{]}+{\text{K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}{\text{][HT}}^{\text{-}}\text{]}={\text{F}}_{{\text{H}}_{\text{2}}\text{T}}\\ {\text{ [Na+]{1+K}}_{{\text{NaT}}^{\text{-}}}{\text{[T}}^{\text{2-}}\text{]}+{\text{K}}_{\text{NaHT}}{\text{[HT}}^{\text{-}}\text{]}\}={\text{F}}_{{\text{H}}_{\text{2}}\text{T}}\\ \text{[Na+]}\text{=}\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{1+{\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[T}}^{\text{2-}}{\text{] + K}}_{\text{NaHT}}{\text{[HT}}^{\text{-}}\text{]}}\end{array}$

Hence, the equation 13-16 is derived from mass balance 13.15.

(b)

Interpretation Introduction

Interpretation:

An expression for ${\text{[T}}^{\text{2-}}\text{]}$ in terms of ${\text{[H}}^{\text{+}}{\text{], [Na}}^{\text{+}}\text{]}$ and various equilibrium constants has to be derived by substituting equilibrium expressions into mass balance equation 13-17

Concept Introduction:

Mass balance equation for ${\text{[H}}_{\text{2}}\text{T]}$:

The mass balance equation for ${\text{[H}}_{\text{2}}\text{T]}$ is:

${\text{F}}_{{\text{H}}_{\text{2}}\text{T}}{\text{ = [H}}_{\text{2}}{\text{T] + [HT}}^{\text{-}}{\text{] + [T}}^{\text{2-}}{\text{] + [NaT}}^{\text{-}}\text{] + [NaHT]}\mathrm{......}\text{(13-17)}$

Answer to Problem 13.17P

The expression for ${\text{[T}}^{\text{2-}}\text{]}$ is:

${\text{[T}}^{\text{2-}}\text{]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\frac{{{\text{[H}}^{\text{+}}\text{]}}^2}{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}+\frac{\text{[H+]}}{{\text{K}}_{\text{2}}}+1+{\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{]+K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}}$

Explanation of Solution

${\text{[H}}_{\text{2}}{\text{T] + [HT}}^{\text{-}}{\text{] + [T}}^{\text{2-}}{\text{] + [NaT}}^{\text{-}}{\text{] + [NaHT] = F}}_{{\text{H}}_{\text{2}}\text{T}}$

Make the following substitutions to find expressions in terms of ${\text{[T}}^{\text{2-}}\text{]}$, ${\text{[H}}^{\text{+}}{\text{] and [Na}}^{\text{+}}\text{]}$:

$\begin{array}{l}{\text{[H}}_{\text{2}}\text{T]=}\frac{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}[{\text{T}}^{\text{2-}}\text{];}{\text{[HT}}^{\text{-}}\text{]=}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}[{\text{T}}^{\text{2-}}\text{];}{\text{[NaT}}^{\text{-}}{\text{]=K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{][T}}^{\text{2-}}\text{]}\\ {\text{[NaHT]=K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}{\text{][HT}}^{\text{-}}{\text{] =K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}[{\text{T}}^{\text{2-}}\text{]}\end{array}$

Put these expressions into the mass balance equation:

$\begin{array}{l}{\text{[H}}_{\text{2}}{\text{T] + [HT}}^{\text{-}}{\text{] + [T}}^{\text{2-}}{\text{] + [NaT}}^{\text{-}}{\text{] + [NaHT] = F}}_{{\text{H}}_{\text{2}}\text{T}}\\ \frac{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}[{\text{T}}^{\text{2-}}\text{]+}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}[{\text{T}}^{\text{2-}}{\text{]+[T}}^{\text{2-}}\text{]}+{\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{][T}}^{\text{2-}}{\text{]+K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}[{\text{T}}^{\text{2-}}{\text{]=F}}_{{\text{H}}_{\text{2}}\text{T}}\end{array}$

Solve for ${\text{[T}}^{\text{2-}}\text{]}$:

${\text{[T}}^{\text{2-}}\text{]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\frac{{{\text{[H}}^{\text{+}}\text{]}}^2}{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}+\frac{\text{[H+]}}{{\text{K}}_{\text{2}}}+1+{\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{]+K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}}$

Hence, the expression for ${\text{[T}}^{\text{2-}}\text{]}$ is obtained.

Conclusion

The expression for ${\text{[T}}^{\text{2-}}\text{]}$ is derived as follows:

${\text{[T}}^{\text{2-}}\text{]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\frac{{{\text{[H}}^{\text{+}}\text{]}}^2}{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}+\frac{\text{[H+]}}{{\text{K}}_{\text{2}}}+1+{\text{K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}{\text{]+K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{2}}}}$

(c)

Interpretation Introduction

Interpretation:

Using the same approach in part (b), the expression for ${\text{[HT}}^{\text{-}}{\text{] and [H}}_{\text{2}}\text{T]}$ has to be derived.

Concept Introduction:

Mass balance equation for ${\text{[H}}_{\text{2}}\text{T]}$:

The mass balance equation for ${\text{[H}}_{\text{2}}\text{T]}$ is:

${\text{F}}_{{\text{H}}_{\text{2}}\text{T}}{\text{ = [H}}_{\text{2}}{\text{T] + [HT}}^{\text{-}}{\text{] + [T}}^{\text{2-}}{\text{] + [NaT}}^{\text{-}}\text{] + [NaHT]}\mathrm{......}\text{(13-17)}$

Answer to Problem 13.17P

The expression for ${\text{[HT}}^{\text{-}}{\text{] and [H}}_{\text{2}}\text{T]}$ is:

$\begin{array}{l}{\text{[HT}}^{\text{-}}\text{]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{1}}}+1+\frac{{\text{K}}_{\text{2}}}{{\text{[H}}^{\text{+}}\text{]}}{\text{+K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{K}}_{\text{2}}}{\text{[H+]}}{\text{+K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}}\\ {\text{[H}}_{\text{2}}\text{T]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{1+\frac{{\text{K}}_1}{{\text{[H}}^{\text{+}}\text{]}}\text{+}\frac{{\text{K}}_1{\text{K}}_2}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}{\text{+K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{K}}_1{\text{K}}_2}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}+{\text{K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{K}}_1}{{\text{[H}}^{\text{+}}\text{]}}}\end{array}$

Explanation of Solution

To find $\left[{\text{HT}}^{\text{-}}\right]$, make the following substitutions in the mass balance for ${\text{H}}_{\text{2}}\text{T}$ :

$\begin{array}{l}\left[{\text{H}}_{\text{2}}\text{T}\right]\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]}{{\text{K}}_{\text{1}}}\left[{\text{HT}}^{\text{-}}\right]\text{;}\left[{\text{T}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{2}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{HT}}^{\text{-}}\right]\text{;}\left[\text{NaHT}\right]{\text{=K}}_{\text{NaHT}}\left[{\text{Na}}^{\text{+}}\right]\left[{\text{HT}}^{\text{-}}\right]\\ \left[{\text{NaT}}^{\text{-}}\right]{\text{=K}}_{{\text{NaT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{2}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{HT}}^{\text{-}}\right]\end{array}$ Then keep these expressions into the mass balance:

$\begin{array}{l}\frac{\left[{\text{H}}^{\text{+}}\right]}{{\text{K}}_{\text{1}}}\left[{\text{HT}}^{\text{-}}\right]\text{+}\left[{\text{HT}}^{\text{-}}\right]\text{+}\frac{{\text{K}}_{\text{2}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{HT}}^{\text{-}}\right]{\text{+K}}_{\text{NaT}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{2}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{HT}}^{\text{-}}\right]{\text{+K}}_{\text{NaT}}\left[{\text{Na}}^{\text{+}}\right]\left[{\text{HT}}^{\text{-}}\right]\\ {\text{=F}}_{{\text{H}}_{\text{2}}\text{T}}\end{array}$

And now solve for $\left[{\text{HT}}^{\text{-}}\right]$

$\left[{\text{HT}}^{\text{-}}\right]\text{=}\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\frac{\left[{\text{H}}^{\text{+}}\right]}{{\text{K}}_{\text{1}}}\text{+1+}\frac{{\text{K}}_{\text{2}}}{\left[{\text{H}}^{\text{+}}\right]}{\text{+K}}_{{\text{NaT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{2}}}{\left[{\text{H}}^{\text{+}}\right]}{\text{+K}}_{\text{NaHT}}\left[{\text{Na}}^{\text{+}}\right]}$

To find $\left[{\text{H}}_{\text{2}}\text{T}\right]$ make the following substitution in the mass balance for ${\text{H}}_{\text{2}}\text{T}$

$\begin{array}{l}\left[{\text{HT}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{H}}_{\text{2}}\text{T}\right]\text{;}\left[{\text{T}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\left[{\text{H}}_{\text{2}}\text{T}\right]\\ \left[{\text{NaT}}^{\text{-}}\right]{\text{=K}}_{{\text{NaT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\left[{\text{T}}^{\text{2-}}\right]{\text{=K}}_{{\text{NaT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\left[{\text{H}}_{\text{2}}\text{T}\right]\\ \left[\text{NaHT}\right]{\text{=K}}_{{\text{NaHT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\left[{\text{HT}}^{\text{-}}\right]{\text{=K}}_{{\text{NaHT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{1}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\left[{\text{H}}_{\text{2}}\text{T}\right]\end{array}$

Then keep these expressions into the mass balance:

$\left[{\text{H}}_{\text{2}}\text{T}\right]\text{+}\frac{{\text{K}}_{\text{1}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{H}}_{\text{2}}\text{T}\right]\text{+}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\left[{\text{H}}_{\text{2}}\text{T}\right]{\text{+K}}_{{\text{NaT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\left[{\text{H}}_{\text{2}}\text{T}\right]{\text{+K}}_{{\text{NaHT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{1}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{H}}_{\text{2}}\text{T}\right]{\text{=F}}_{{\text{H}}_{\text{2}}\text{T}}$

And solve for $\left[{\text{H}}_{\text{2}}\text{T}\right]$

$\left[{\text{H}}_{\text{2}}\text{T}\right]\text{=}\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\text{1+}\frac{{\text{K}}_{\text{1}}}{\left[{\text{H}}^{\text{+}}\right]}\text{+}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{\text{+K}}_{{\text{NaT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{\text{+K}}_{{\text{NaHT}}^{\text{-}}}\left[{\text{Na}}^{\text{+}}\right]\frac{{\text{K}}_{\text{1}}}{\left[{\text{H}}^{\text{+}}\right]}}$

Conclusion

The expression for ${\text{[HT}}^{\text{-}}{\text{] and [H}}_{\text{2}}\text{T]}$ is derived as follows:

$\begin{array}{l}{\text{[HT}}^{\text{-}}\text{]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{K}}_{\text{1}}}+1+\frac{{\text{K}}_{\text{2}}}{{\text{[H}}^{\text{+}}\text{]}}{\text{+K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{K}}_{\text{2}}}{\text{[H+]}}{\text{+K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}}\\ {\text{[H}}_{\text{2}}\text{T]}=\frac{{\text{F}}_{{\text{H}}_{\text{2}}\text{T}}}{1+\frac{{\text{K}}_1}{{\text{[H}}^{\text{+}}\text{]}}\text{+}\frac{{\text{K}}_1{\text{K}}_2}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}{\text{+K}}_{{\text{NaT}}^{\text{-}}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{K}}_1{\text{K}}_2}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}+{\text{K}}_{\text{NaHT}}{\text{[Na}}^{\text{+}}\text{]}\frac{{\text{K}}_1}{{\text{[H}}^{\text{+}}\text{]}}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Using excel spreadsheet, the pH of the given solution has to be computed.

Answer to Problem 13.17P

The pH of the given solution is $4.264$

Explanation of Solution

The spreadsheet used to compute pH of the given solution is shown in the below figure 1.

Figure 1

From the above spreadsheet, the pH of the solution is found as $4.264$

Conclusion

The pH of the given solution is found out as $4.264$