Chapter 16, Problem 16.AE

Interpretation Introduction

Interpretation:

The potential at the given volumes of ${\text{Ce}}^{\text{4+}}$ has to be calculated and the titration curve has to be drawn.

Answer to Problem 16.AE

The potential for given volume of ${\text{Ce}}^{\text{4+}}$ are,

At $\text{0}\text{.100}\text{mL}$: $\text{- 0}\text{.161V}$

At $\text{1}\text{.00}\text{mL}$: $\text{-0}\text{.130 V}$

At $\text{5}\text{.00}\text{mL}$: $\text{-0}\text{.102 V}$

At $\text{9}\text{.50}\text{mL}$: $\text{-0}\text{.064 V}$

At $\text{10}\text{.0}\text{mL}$: $\text{0}\text{.342 V}$

At $\text{10}\text{.10}\text{mL}$: $\text{1}\text{.11 V}$

At $\text{12}\text{.0}\text{mL}$: $\text{1}\text{.19 V}$

The titration curve is,

Figure 1

Explanation of Solution

Given

Amount of solution of ${\text{Sn}}^{\text{2+}}$= $\text{20}\text{.0 mL}$

Strength of ${\text{Sn}}^{\text{2+}}$ solution = $\text{0}\text{.00500 M}$

Strength of ${\text{Ce}}^{\text{4+}}$ solution = $\text{0}\text{.0200 M}$

To explain: The potential for given volume of ${\text{Ce}}^{\text{4+}}$

$\text{Titration:}{\text{Sn}}^{\text{2+}}{\text{+2Ce}}^{\text{4+}}\stackrel{}{\to }{\text{Sn}}^{\text{4+}}{\text{+2Ce}}^{\text{3+}}{\text{V}}_{\text{e}}\text{=10}\text{.0mL}$

At $\text{0}\text{.100}\text{mL}$:

${\text{V}}_{\text{e}}\text{=0}\text{.100mL/10}\text{.00mL}$

$\begin{array}{l}{\text{E}}_{\text{+}}\text{= 0}\text{.139-}\frac{\text{0}\text{.5916}}{\text{2}}\text{log}\frac{\left[{\text{Sn}}^{\text{2+}}\right]}{\left[{\text{Sn}}^{\text{4+}}\right]}\\ \\ \text{= 0}\text{.139-}\frac{\text{0}\text{.5916}}{\text{2}}\text{log}\frac{\text{9}\text{.90}}{\text{0}\text{.100}}\\ \\ \text{= 0}\text{.080}\text{V}\\ \\ {\text{E = E}}_{\text{+}}{\text{-E}}_{\text{-}}\text{= 0}\text{.080-0}\text{.241= - 0}\text{.161}\text{V}\end{array}$

At $\text{10}\text{.0}\text{mL}$:

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{2E}}_{\text{+}}\text{=2}\left(\text{0}\text{.139}\right)\text{- 0}\text{.05916 log}\frac{\left[{\text{Sn}}^{\text{2+}}\right]}{\left[{\text{Sn}}^{\text{4+}}\right]}\\ \\ {\text{E}}_{\text{+}}\text{=1}\text{.47- 0}\text{.05916 log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\end{array}}\\ {\text{3E}}_{\text{+}}\text{=1}\text{.748 - 0}\text{.05916 log }\frac{\left[{\text{Sn}}^{\text{2+}}\right]\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Sn}}^{\text{4+}}\right]\left[{\text{Ce}}^{\text{4+}}\right]}\end{array}$

At the equivalence point,

$\left[{\text{Sn}}^{\text{4+}}\right]\text{ =}\frac{\text{1}}{\text{2}}\left[{\text{Ce}}^{\text{3+}}\right]$ and $\left[{\text{Sn}}^{\text{2+}}\right]$= $\frac{\text{1}}{\text{2}}\left[{\text{Ce}}^{\text{4+}}\right]$, which forms the log term zero, So ${\text{3E}}_{\text{+}}\text{=1}\text{.748 }$ and ${\text{E}}_{\text{+}}\text{= 0}\text{.583}\text{V }$

$\begin{array}{l}{\text{E=E}}_{\text{+}}{\text{-E}}_{\text{-}}\text{= 0}\text{.583 - 0}\text{.241}\\ \\ \text{= 0}\text{.342}\text{V}\end{array}$

At $\text{10}\text{.10 mL}$: The first $\text{10}\text{.10 mL}$ used to making ${\text{Ce}}^{\text{3+}}$. Next $\text{0}\text{.10 mL}$ remains unreacted ${\text{Ce}}^{\text{4+}}$.

$\begin{array}{l}{\text{E}}_{\text{+}}\text{= 1}\text{.47- 0}\text{.5916 log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\\ \\ \text{= 1}\text{.47- 0}\text{.5916}\text{log}\frac{10.0}{\text{0}\text{.10}}\\ \\ \text{= 1}\text{.352}\text{V}\\ \\ {\text{E = E}}_{\text{+}}{\text{- E}}_{\text{-}}\text{= 1}\text{.352 - 0}\text{.241= - 0}\text{.11}\text{V}\end{array}$

To write: titration curve

$mL$                       E $\left(V\right)$
0.100                 -0.161
1.00                   -0.130
5.00                   -0.102
9.50                   -0.064
10.00                0.342
10.10                 1.11
12.00                 1.19

Figure 1