Question

The sulfur content of insoluble sulfides that do not readily dissolve in acid can be measured by oxidation with Br₂ to .²⁵ Metal ions are then replaced with H⁺ by an ion-exchange column, and sulfate is precipitated as BaSO₄ with a known excess of BaCl₂. The excess Ba²⁺ is then titrated with EDTA to determine how much was present. (To make the indicator end point clearer, a small, known quantity of Zn²⁺ also is added. The EDTA titrates both the Ba²⁺ and the Zn²⁺.) Knowing the excess Ba²⁺, we can calculate how much sulfur was in the original material. To analyze the mineral sphalerite (ZnS, FM 97.46), 5.89 mg of powdered solid were suspended in a mixture of CCl₄ and H₂O containing 1.5 mmol Br₂. After 1 h at 20⁰ C and 2 h at 50⁰ C, the powder dissolved and the solvent and excess Br₂ were removed by heating. The residue was dissolved in 3 mL of water and passed through an ion-exchange column to replace Zn²⁺ with H⁺. Then 5.000 mL of 0.014 63 M BaCl₂ were added to precipitate all sulfate as BaSO₄. After the addition of 1.000 mL of 0.010 00 M ZnCl₂ and 3 mL of ammonia buffer, pH 10, the excess Ba²⁺ and Zn²⁺ required 2.39 mL of 0.009 63 M EDTA to reach the Calmagite end point. Find the weight percent of sulfur in the sphalerite. What is the theoretical value?

Short answer

The weight percent of sulfur in sphalerite is 32.76%. The theoretical weight percent of sulfur in pure ZnS will be 33.6%

Step-by-step solution

Given Information

The sulfur content of insoluble sulfides generally measured by oxidation with Br₂ to ${\mathrm{SO}}_4^{2-}$. An ion exchange column must be used to replace metal ions with H⁺. After precipitation of sulfate as BaSO₄, the excess Ba²⁺ ion is titrated with EDTA.

Amount of powdered solid taken = 5.89 mg

Amount of BaCl₂ added for precipitation of BaSO₄ = 5.0 mL of 0.01463 M BaCl₂

Amount of ZnCl₂ added = 1.0 mL of 0.01M ZnCl₂

Amount of ammonia(pH=10) added = 3 mL

EDTA required to attain Calmagite end point =0.00963 M

Determine the amount of sulfur present

Total amount of standard Ba²⁺ and Zn²⁺ added to sulfate solution

$$\begin{gathered} =\left(5\mathrm{mL}\right)(0.01463{\mathrm{BaCl}}_2)+\left(1\mathrm{mL}\right)(0.01{\mathrm{ZnCl}}_2) \\ =0.0832\mathrm{mmol} \end{gathered}$$

Amount of EDTA required for titration of excess Ba²⁺ ion

$$\begin{gathered} =(2.39\mathrm{mL})(0.00963\mathrm{M}) \\ =0.023\mathrm{mmol} \end{gathered}$$

Therefore, the original amount of sulfur in the solid

$$\begin{gathered} =(0.0832-0.023)\mathrm{mmol} \\ =0.0602\mathrm{mmol} \end{gathered}$$

Molecular weight of sulfur = 32.066 mg/mmol

Therefore, mass of sulfur

$$\begin{gathered} =(0.0602\mathrm{mmol})(32.066\mathrm{mg}/\mathrm{mmol}) \\ =1.93\mathrm{mg} \end{gathered}$$

Determine the theoretical value of sulfur present

Weight percentage of sulfur in the solid (spharelite) taken

$$\begin{gathered} =\frac{1.93\mathrm{mg}}{5.89\mathrm{mg}}\times 100 \\ =32.76\% \end{gathered}$$

Theoretical weight percent of sulfur in pure ZnS

$$\begin{gathered} =\frac{32.76\mathrm{g}\mathrm{S}}{97.46\mathrm{gZnS}} \\ =33.6\% \end{gathered}$$

Therefore, theoretical weight percent of sulfur in pure ZnS will be 33.6%