${\mathrm{H}}_2\mathrm{S}$was measured by slowly adding 25.00mLof aqueous ${\mathrm{H}}_2\mathrm{S}$ to 25.00mL of acidified standard$0.01044{\mathrm{Ml}}_3^{-}$to precipitate elemental sulfur. (If $\left[{\mathrm{H}}_2\mathrm{S}\right]>0.01\mathrm{M}$, then precipitated sulfur traps some${\mathrm{l}}_3^{-}$solution, which is not subsequently titrated.) The remaining${\mathrm{l}}_3^{-}$was titrated with 14.44mL of $0.009336{\mathrm{MNa}}_2{\mathrm{S}}_2{\mathrm{O}}_3$. Find the molarity of the${\mathrm{H}}_2\mathrm{S}$solution. Should starch indicator be added to this titration at the beginning or just before the end point?

Molarity is also known as amount concentration, molar concentration, or substance concentration. It is a measure of the concentration of a chemical species, specifically a solute, in a solution. It is a substance per unit volume of solution. The most common unit for the term molarity in chemistry is the number of moles per litre.

molarity = moles of solute / volume of solution in litres

$\mathrm{i}.\mathrm{e}.\mathrm{M}=\frac{\mathrm{n}}{\mathrm{V}}$

First we can write two half-reactions for the reaction of ${\mathrm{H}}_2\mathrm{S}$and triiodide:

Reduction $:{\mathrm{l}}_3^{-}+2{\mathrm{e}}^{-}\rightleftharpoons 3{\mathrm{l}}^{-}$

Oxidation $:{\mathrm{H}}_2\mathrm{S}\rightleftharpoons \mathrm{S}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

Since the number of electrons are equal in both reactions, we can just sum them:

${\mathrm{l}}_3^{-}+2\mathrm{e}+{\mathrm{H}}_2\mathrm{S}\rightleftharpoons 3{\mathrm{l}}^{-}+\mathrm{S}+2{\mathrm{H}}^{+}+2-\mathrm{e}$

The net reaction is:

${\mathrm{l}}_3^{-}+{\mathrm{H}}_2\mathrm{S}\rightleftharpoons 3{\mathrm{l}}^{-}+\mathrm{S}+2{\mathrm{H}}^{+}$

Since the number of electrons are equal in both reactions, we can just sum them:

${\mathrm{l}}_3^{-}+2{\mathrm{e}}^{-}+{\mathrm{H}}_2\mathrm{S}\rightleftharpoons 3{\mathrm{l}}^{-}+\mathrm{S}+2{\mathrm{H}}^{+}$

The net reaction is:

${\mathrm{l}}_3^{-}+2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\rightleftharpoons 3{\mathrm{l}}^{-}+{\mathrm{S}}_4{\mathrm{O}}_6^{2-}$

The excess of triiodide was titrated with sodium thiosulfate . We can write the reaction so we can determine the excess moles of triiodide:

role="math" localid="1654851645784" ${\mathrm{l}}_3^{-}+2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\rightleftharpoons 3{\mathrm{l}}^{-}+{\mathrm{S}}_4{\mathrm{O}}_6^{2-}$

The excess of triiodide was titrated with sodium thiosulfate. We can write the reaction so we can determine the excess moles of triiodide:

${\mathrm{l}}_3^{-}+2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\rightleftharpoons 3{\mathrm{l}}^{-}+{\mathrm{S}}_4{\mathrm{O}}_6^{2-}$

Since we have a iodometric titration in the task,starch should be added right before the equivalence point so that the triiodide doesn’t bind to it during the reaction.

Therefore the weight percent of Cu in the salt is 11.45%.