Question

17-18. ${\mathrm{H}}_2\mathrm{S}\left(\mathrm{aq}\right)$ can be analysed by titration with coulometric ally generated ${\mathrm{l}}_2.$

role="math" localid="1654768790950" ${\mathrm{H}}_2\mathrm{S}+{\mathrm{l}}_2\to \mathrm{S}\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}+2{\mathrm{l}}^{-}$

To $50.00\mathrm{mL}$ of sample were added 4gof KI. Electrolysis required812s at. 52.6mA . Calculate the concentration ofrole="math" localid="1654769975824" ${\mathrm{H}}_2\mathrm{S}(\mathrm{\mu g}/\mathrm{mL})$in the sample.

Short answer

The concentration of${\mathrm{H}}_2\mathrm{S}(\mathrm{\mu g}/\mathrm{mL})$ in the sample is$\mathrm{\gamma }=151\mathrm{\mu g}/\mathrm{mL}$

Step-by-step solution

Moles of iodine:

One mole (abbreviated mol) is equal to $6.022\times {10}^{23}$tiny particles of the material in question. As a result, $6.022\times {10}^{23}$atoms are referred to as 1 mol l. The $7\times {10}^{17}$atoms of iodine in the RDA we've been talking about would be $(7\times {10}^{17})/(6.022\times {10}^{23})$mol Br or 0.0000012 mol iodine.

Calculate mols of iodine:

To see how many electrons are used we need to write the equation for the reduction of iodine :

$I_2+2e^{-}↤\Rightarrow 2I^{-}$

To calculate mols of iodine used we can use the following relation:

$\mathrm{n}\left({\mathrm{I}}_2\right)=\frac{\mathrm{I}.\mathrm{t}}{\mathrm{n}.\mathrm{F}}$

whereis the Faraday constant, F= 96485C/mol

By inserting the known data, we get:

$\mathrm{n}\left({\mathrm{l}}_2\right)=\frac{52.6\times {10}^{-3}\mathrm{A}.812\mathrm{s}}{2.96485\mathrm{C}/\mathrm{mol}}=2.213\times {10}^{-4}\mathrm{mol}$

Calculate the concentration of  H2S(μg/mL)in the sample

We can see from the equation in the task question that the mols of iodine are equal to mols of ${\mathrm{H}}_2\mathrm{S}$so the mass concentration ${\mathrm{H}}_2\mathrm{S}$is as follows:

$$\begin{gathered} \mathrm{\gamma }\left({\mathrm{H}}_2\mathrm{S}\right)=\frac{\mathrm{m}\left({\mathrm{H}}_2\mathrm{S}\right)}{\mathrm{V}\left(\mathrm{solution}\right)}=\frac{\mathrm{n}\left({\mathrm{H}}_2\mathrm{S}\right).\mathrm{M}\left({\mathrm{H}}_2\mathrm{S}\right)}{\mathrm{V}\left(\mathrm{solution}\right)} \\ \mathrm{\gamma }\left({\mathrm{H}}_2\mathrm{S}\right)=\frac{2.213\times {10}^{-10}\mathrm{mol}.34.08\mathrm{g}/\mathrm{mol}}{50\mathrm{mL}}=1.51\times {10}^{-4}\mathrm{g}/\mathrm{ml}=151\mathrm{\mu g}/\mathrm{ml} \end{gathered}$$