Question

What fraction of ethane-1,2-dithiol is in each form (H2A,HA2, A22) at pH 8.00? at pH 10.00?

Short answer

The answer is;

For ${\mathrm{H}}_2\mathrm{A}$;

$\mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)$(At pH=8.00), $\mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)$(at pH=10.00)

For $\mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)$;

$\mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)$(At pH=8.00), $\mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)$ (at pH=10.00)

For $\mathrm{\alpha }\left({\mathrm{A}}^{2-}\right)$;

$\mathrm{\alpha }\left({\mathrm{A}}^{2-}\right)$(At pH=8.00), $\mathrm{\alpha }\left(A^{2-}\right)$ (at pH=10.00)

Step-by-step solution

Step 1: Ethane-1,2-dithiol

EDT, commonly known as ethane-1,2-dithiol, is a colourless liquid possessing the chemical formula C₂H₄(SH)₂. It has a highly distinctive stench that many people liken to that of rotting cabbage. It is a typical building component in the synthesis of organic compounds and a superior ligand for metal ions.

Calculating the fraction of ethane-1,2-dithiol in its different forms and at different pH

We will use the following:

$$\begin{gathered} K_{1,2..}={10}^{-p{K}_{a1,2...}} \\ p{K}_{a1}=8.85\to K_1={10}^{-8.85},p{K}_{a2}=10.43\to K_2={10}^{-10.43} \end{gathered}$$

The fraction of ${\mathrm{H}}_{2}A$,

a. at pH=8.00

$$\begin{gathered} \alpha \left(H_{2}A\right)=\frac{\left[H^{+}\right]}{[H^{+}{]}^2+[H^{+}]K_1+K_1{K}_2} \\ \alpha \left(H_{2}A\right)=\frac{({10}^{-8}{)}^2}{{\left({10}^{-8}\right)}^2+\left(\right({10}^{-8})\times {10}^{-8.85})+({10}^{-8.85}\times {10}^{-10.43})} \\ \alpha \left(H_{2}A\right)=0.876 \end{gathered}$$

b. at pH=10.00

role="math" localid="1663564741770" $$\begin{gathered} \mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)=\frac{\left[{\mathrm{H}}^{+}\right]}{[{\mathrm{H}}^{+}{]}^2+[{\mathrm{H}}^{+}]{\mathrm{K}}_1+{\mathrm{K}}_1{\mathrm{K}}_2} \\ \mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)=\frac{{\left({10}^{-8}\right)}^2}{{\left({10}^{-10}\right)}^2+\left(\right({10}^{-10})\times {10}^{-8.85})+({10}^{-8.85}\times {10}^{-10.43})} \\ \mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)=0.491 \end{gathered}$$

The fraction of ${\mathrm{HA}}^{-}$,

a. at pH=8.00

$$\begin{gathered} \mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)=\frac{{\mathrm{K}}_1\left[{\mathrm{H}}^{+}\right]}{[{\mathrm{H}}^{+}{]}^2+[{\mathrm{H}}^{+}]{\mathrm{K}}_1+{\mathrm{K}}_1{\mathrm{K}}_2} \\ \mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)=\frac{({10}^{-8})\times {10}^{-8.85}}{{\left({10}^{-8}\right)}^2+\left(\right({10}^{-8})\times {10}^{-8.85})+({10}^{-8.85}\times {10}^{-10.43})} \\ \mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)=0.123 \end{gathered}$$

b. at pH=10.00

$$\begin{gathered} \mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)=\frac{{\mathrm{K}}_1\left[{\mathrm{H}}^{+}\right]}{[{\mathrm{H}}^{+}{]}^2+[{\mathrm{H}}^{+}]{\mathrm{K}}_1+{\mathrm{K}}_1{\mathrm{K}}_2} \\ \mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)=\frac{({10}^{-10})\times {10}^{-8.85}}{{\left({10}^{-10}\right)}^2+\left(\right({10}^{-10})\times {10}^{-8.85})+({10}^{-8.85}\times {10}^{-10.43})} \\ \mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)=0.694 \end{gathered}$$

The fraction of $A^{2-}$,

a. at pH=8.00

$$\begin{gathered} \mathrm{\alpha }\left(A^{2-}\right)=\frac{{\mathrm{K}}_1{\mathrm{K}}_2}{[{\mathrm{H}}^{+}{]}^2+[{\mathrm{H}}^{+}]{\mathrm{K}}_1+{\mathrm{K}}_1{\mathrm{K}}_2} \\ \mathrm{\alpha }\left(A^{2-}\right)=\frac{({10}^{-10.43})\times {10}^{-8.85}}{{\left({10}^{-8}\right)}^2+\left(\right({10}^{-8})\times {10}^{-8.85})+({10}^{-8.85}\times {10}^{-10.43})} \\ \mathrm{\alpha }\left(A^{2-}\right)=4.54\times {10}^{-4} \end{gathered}$$

b. at pH=10.00

$$\begin{gathered} \mathrm{\alpha }\left(A^{2-}\right)=\frac{{\mathrm{K}}_1{\mathrm{K}}_2}{[{\mathrm{H}}^{+}{]}^2+[{\mathrm{H}}^{+}]{\mathrm{K}}_1+{\mathrm{K}}_1{\mathrm{K}}_2} \\ \mathrm{\alpha }\left(A^{2-}\right)=\frac{({10}^{-10.43})\times {10}^{-8.85}}{{\left({10}^{-10}\right)}^2+\left(\right({10}^{-10})\times {10}^{-8.85})+({10}^{-8.85}\times {10}^{-10.43})} \\ \mathrm{\alpha }\left(A^{2-}\right)=0.257 \end{gathered}$$

The final answer is;

For $\mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)$;

$\mathrm{\alpha }\left({\mathrm{H}}_2\mathrm{A}\right)$ (At pH=8.00), (at pH=10.00)

For $\mathrm{\alpha }\left({\mathrm{HA}}^{-}\right)$;

$\mathrm{\alpha }\left(H{A}^{-}\right)$(At pH=8.00), (at pH=10.00)

For $\mathrm{\alpha }\left(A^{2-}\right)$;

$\mathrm{\alpha }\left(A^{2-}\right)$ (At pH=8.00), (at pH=10.00)