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Chapter 16: Q22P (page 392)

Why is iodine almost always used in a solution containing excess $l^{-}$ ?

Short Answer

Expert verified

Iodine is always used in a solution with excess $l^{-}$ because pure $l_{2}$ is not a polar substance and it cannot be dissolved in water.

Step by step solution

01

Properties of Iodine.

Iodine is a chemical element with atomic number 53 and the symbol l .
At typical conditions, it exists as a semi-lustrous, non-metallic solid that melts to produce a deep violet liquid at $114 °$ C and boils to form a violet gas at $184 ° C$
Iodine is found in a variety of oxidation states, including iodide $(I^{-})$ , iodate $({IO}_{3}^{-})$ , and periodate anions.
It is the heaviest nutrient in terms of necessary minerals.

02

To determine the iodine always used in solution containing

lodine is always used in a solution with excess $I^{-}$ because pure $I_{2}$ is not a polar substance and it cannot be dissolved in water.
Because of that, iodide solution(for example, Kl) must be added to iodine to increase its solubility,
The following reactions occurs:
$I_{2} (s) + I^{-} (aq) ⇌ I_{3}^{-} (aq)$
In this reaction triodide is formed by a complexation reaction between iodide and iodine and the solubility of $I_{2}$ is increased.

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Most popular questions from this chapter

Whenof unknown were passed through a Jones reductor, molybdate ion $({MoO}_{4}^{2 -})$ was converted into ${Mo}^{3 +}$ . The filtrate required 16.43mLof $0.01033 {MKMnO}_{4}$ to reach the purple end point.

${MnO}_{4}^{-} + {Mo}^{3 +} \to {Mn}^{2 +} + {MoO}_{2}^{2 +}$

A blank required 0.04mL. Balance the reaction and find the molarity of Molybdate in the unknown.

Consider the titration of 25.0 mLof $0.0500 M S n^{2 +}$ with $0.100 M F e^{3 +}$ in 1MHCI to give $F e^{2 +}$ and $S n^{4 +}$ , using Pt and calomel electrodes.

(a) Write a balanced titration reaction.

(b) Write two half-reactions for the indicator electrode.

(c) Write two Nernst equations for the cell voltage.

(d) Calculate Eat the following volumes of $F e^{3} : 1.0, 12.5, 24.0, 25.0, 26.0$ and 30.0 mL. Sketch the titration curve.

Write balanced half-reactions in which $M n O_{4}^{-}$ acts as an oxidant at

$(a) pH = 0; (b) pH = 10; (c) pH = 15 .$

Compute the titration curve for Demonstration 16-1, in which 400.0mLof $37.5 {mMFe}^{2 +}$ are titrated with $20.0 {mMMnO}_{4}^{-}$ at a fixed pH of 0.00 in $1 {MH}_{2} {SO}_{4}$ . Calculate the potential versus S.C.E. at titrant volumes of 1.0,7.5,14.0,15.0,16.0, and 30.0 mL and sketch the titration curve.

Aqueous glycerol solution weighing 100.0m gwas treated with 50.0 mL of 0.083 7 M ${Ce}^{4 +}$ in 4 MHCI $O_{4}$ at $60 °$ for15minto oxidize glycerol to formic acid.

${CH}_{2} - CH - {CH}_{2} | | | OH OH OH$ ${HCO}_{2} H$

Glycerol Formic acid

FM92.095

The excess ${Ce}^{4 +}$ required 12.11mL of 0.044 8 MF $e^{2 +}$ to reach a ferroin end point. Find wt%glycerol in the unknown.

See all solutions

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