1. Fractional composition in a tetraprotic system. Prepare a fractional composition diagram analogous to Figure 10-4 for the tetraprotic system derived from hydrolysis of ${}^{{\mathrm{Cr}}^{+}}$:

localid="1654853037629" $$\begin{gathered} C{r}^{3+}+H_{2}O\leftrightharpoons Cr{\left(OH\right)}^{2+}+H^{+}{K}_{a1}={10}^{-3.80} \\ Cr{\left(OH\right)}^{2+}+H_{2}O\leftrightharpoons Cr{\left(OH\right)}_2^{+}+H^{+}{K}_{a2}={10}^{-6.40} \\ Cr{\left(OH\right)}_2^{+}+H_{2}O\leftrightharpoons Cr{\left(OH\right)}_3\left(aq\right)+H^{+}{K}_{a3}={10}^{-6.40} \\ Cr{\left(OH\right)}_3\left(aq\right)+H_{2}O\leftrightharpoons Cr{\left(OH\right)}_4^{-}+H^{+}{K}_{a4}{=}^{-11.40} \end{gathered}$$

(Yes, the values oflocalid="1654853051658" ${}^{K_{a2}}$andlocalid="1654853058271" ${}^{K_{a3}}$are equal.)

(a) Use these equilibrium constants to prepare a fractional composition diagram for this tetraprotic system.

(b) You should do this part with your head and your calculator, not your spreadsheet. The solubility of is given by

localid="1654853063951" $Cr{\left(OH\right)}_3\left(S\right)Cr{\left(OH\right)}_3\left(aq\right)K_{a3}={10}^{-6.80}$

What concentration of localid="1654853075036" ${}^{\mathrm{Cr}{\left(\mathrm{OH}\right)}_3\left(\mathrm{aq}\right)}$is in equilibrium with solid localid="1654853085944" ${}^{Cr{\left(OH\right)}_3\left(aq\right)\left(S\right)}$?

(c) What concentration oflocalid="1654853094735" ${}^{Cr{\left(OH\right)}^{2+}}$is in equilibrium with localid="1654853101499" ${}^{Cr{\left(OH\right)}_3\left(aq\right)\left(S\right)}$if the solution localid="1654853109266" ${}^{pH}$is adjusted tolocalid="1654853117749" ${}^{4.00}$?

Our goal is to find an expression for the fraction of an acid in each form (HA and ${}^{{\mathbf{A}}^{\mathbf{-}}\mathbf{}\mathbf{)}}$as a function of ${}^{\mathbf{pH}\mathbf{.}}$We can do this by combining the equilibrium constant with the mass balance. Consider an acid with formal concentration F:

$\mathbf{HA}\mathbf{}\mathbf{\leftrightharpoons }{\mathbf{H}}^{\mathbf{+}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}{\mathbf{A}}^{\mathbf{-}}$

$$\begin{gathered} {\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\left[\mathrm{HA}\right]} \\ \mathbf{}\mathbf{Massbalance}\mathbf{}\mathbf{:}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{\left[}\mathbf{HA}\mathbf{\right]}\mathbf{+}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]} \end{gathered}$$

(a)

The spreadsheets for the given cells.

The values for cells D2, E2, F2, G2, H2 and I2 are calculated via:

$$\begin{gathered} \mathbf{D}\mathbf{2}\mathbf{=}\mathbf{C}{\mathbf{2}}^{\mathbf{4}\mathbf{+}}\mathbf{}\mathbf{A}\mathbf{2}\mathbf{C}\mathbf{4}\mathbf{}\mathbf{+}\mathbf{}\mathbf{A}\mathbf{2}\mathbf{A}\mathbf{4}\mathbf{C}{\mathbf{2}}^{\mathbf{2}}\mathbf{}\mathbf{*}\mathbf{}\mathbf{A}\mathbf{2}\mathbf{A}{\mathbf{4}}^{\stackrel{\mathbf{°}}{\mathbf{a}}}\mathbf{}\mathbf{A}{\mathbf{6}}^{\mathbf{*}}\mathbf{A}\mathbf{8} \\ \mathbf{E}\mathbf{2}\mathbf{=}\mathbf{C}{\mathbf{2}}^{\mathbf{4}}\mathbf{/}\mathbf{D}\mathbf{2} \\ \mathbf{F}\mathbf{2}\mathbf{=}\mathbf{A}{\mathbf{2}}^{\stackrel{\mathbf{°}}{\mathbf{a}}}\mathbf{}\mathbf{C}{\mathbf{2}}^{\mathbf{3}}\mathbf{/}\mathbf{D}\mathbf{2} \\ \mathbf{G}\mathbf{2}\mathbf{=}\mathbf{A}{\mathbf{2}}^{\stackrel{\mathbf{°}}{\mathbf{a}}}\mathbf{}\mathbf{A}{\mathbf{4}}^{\mathbf{*}}\mathbf{C}{\mathbf{2}}^{\mathbf{2}}\mathbf{/}\mathbf{D}\mathbf{2} \\ \mathbf{H}\mathbf{2}\mathbf{=}\mathbf{A}{\mathbf{2}}^{\stackrel{\mathbf{°}}{\mathbf{a}}}\mathbf{}\mathbf{A}{\mathbf{4}}^{\mathbf{*}}\mathbf{A}{\mathbf{6}}^{\mathbf{*}}\mathbf{C}\mathbf{2}\mathbf{/}\mathbf{D}\mathbf{2} \\ \mathbf{I}\mathbf{2}\mathbf{=}\mathbf{A}{\mathbf{2}}^{\stackrel{\mathbf{°}}{\mathbf{a}}}\mathbf{}\mathbf{A}{\mathbf{4}}^{\mathbf{*}}\mathbf{A}{\mathbf{6}}^{\stackrel{\mathbf{°}}{\mathbf{a}}}\mathbf{}\mathbf{A}\mathbf{8}\mathbf{/}\mathbf{D}\mathbf{2} \end{gathered}$$

b) Calculating the :concentration of ${}^{\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}}$which is in equilibrium with solid ${}^{\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathbf{:}}$

$$\begin{gathered} \mathbf{K}\mathbf{}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{84}}\mathbf{=}{\mathbf{\left[}\mathbf{Cr}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\mathbf{\right]}}_{\mathbf{aq}} \\ {\mathbf{\left[}\mathbf{Cr}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\mathbf{\right]}}_{\mathbf{aq}}\mathbf{}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{84}} \\ {\mathbf{\left[}\mathbf{Cr}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\mathbf{\right]}}_{\mathbf{aq}}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{M} \end{gathered}$$

Therefore the concentration of ${}^{\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}}$is ${}^{\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{M}}$.

(c)

pH=4.00

First we will calculate the concentration of${}^{\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}^{\mathbf{+}}}$

${\mathbf{K}}_{\mathbf{a}\mathbf{3}}\mathbf{=}\mathbf{}\frac{{\mathbf{\left[}\mathbf{Cr}\left(\mathrm{OH}\right)\mathbf{\right]}}_{\mathbf{3}}\mathbf{}{\mathbf{H}}^{\mathbf{+}}\mathbf{]}}{\left[\mathrm{Cr}{\left(\mathrm{OH}\right)}_2^{+}\right]}$

${\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{84}}\mathbf{}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{84}}\mathbf{\right]}\mathbf{\left[}{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}}\mathbf{\right]}}{\left[\mathrm{Cr}{\left(\mathrm{OH}\right)}_2^{+}\right]}$

$\mathbf{\left[}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}^{\mathbf{+}}\mathbf{\right]}\mathbf{}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{44}}$

Next we will calculate the concentration of ${}^{\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}^{\mathbf{+}}}$:

${\mathbf{K}}_{\mathbf{a}\mathbf{2}}\mathbf{=}\mathbf{}\frac{\mathbf{\left[}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}^{\mathbf{+}}\mathbf{\right]}\mathbf{}\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}}{\left[\mathrm{Cr}{\left(\mathrm{OH}\right)}^{2+}\right]}$

${\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{40}}\mathbf{}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{44}}\mathbf{\right]}\mathbf{\left[}{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}}\mathbf{\right]}}{\left[\mathrm{Cr}{\left(\mathrm{OH}\right)}^{2+}\right]}$

$\mathbf{\left[}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{04}}\mathbf{M}$

Therefore the concentration of localid="1654853561905" ${}^{\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}^{\mathbf{2}\mathbf{+}}}$is .