.1724g of a mineral containing MnO₂ was dissolved and then treated with excess iodide according to the following unbalanced reaction: MnO2(s) + H+ → Mn²+ + I2 + H₂O The I₂ released was titrated with a solution of Na₂S2O3 0.07320 mol/L, requiring 14.65 mL to reach the point end of degree
.1724g of a mineral containing MnO₂ was dissolved and then treated with excess iodide according to the following unbalanced reaction: MnO2(s) + H+ → Mn²+ + I2 + H₂O The I₂ released was titrated with a solution of Na₂S2O3 0.07320 mol/L, requiring 14.65 mL to reach the point end of degree
Chemistry by OpenStax (2015-05-04)
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Chapter19: Transition Metals And Coordination Chemistry
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0.1724g of a mineral containing MnO₂ was dissolved and then treated with excess iodide according to the following unbalanced reaction:
MnO2(s) + H+ → Mn²+ + I2 + H₂O
The I₂ released was titrated with a solution of Na₂S2O3 0.07320 mol/L, requiring 14.65 mL to reach the point end of degree
a) Determine the percentage of MnO₂ in the ore.
b) When iodine solutions are used as titrants these are prepared by dissolving 1₂ in concentrated KI. Explain the reason for the need for KI in the preparation of these solutions?
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