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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

Consider the titration of 25.0mL of 0.0500MSn²⁺with 0.100MFe³⁺in 1MHCI to give Fe²⁺ and Sn⁴⁺, 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) Calculateat E the following volumes of Fe³⁺: 1.0,12.5,24.0,25.0,26.0 and 30.0mL. Sketch the titration curve.

Calcium fluorapatite $({Ca}_{10} ({PO}_{4})_{6} F_{2}, FM 1008.6)$ laser crystals were doped with chromium to improve their efficiency. It was suspected that the chromium could be in the $+ 4$ oxidation state.

To measure the total oxidizing power of chromium in the material, a crystal was dissolved in $2.9 M H C L O_{4}$ at $100 ° C$ , cooled to $20 ° C$ , and titrated with standard $F e^{2 +}$ , using Pt and Ag - AgCl electrodes to find the end point. Chromium above the 3 + state should oxidize an equivalent amount of $F e^{2 +}$ in this step. That is, $C r^{4 +}$ would consume one $F e^{2 +}$ , and $C r^{6 +}$ in $C r_{2} O_{7}^{2 -}$ would consume three $F e^{2 +}$ :

role="math" localid="1664873864085" $C r^{4 +} + F e^{2 +} \to C r^{3 +} + F e^{3 +} \frac{1}{2} C r_{2} O_{7}^{2 -} + 3 F e^{2 +} \to C r^{3 +} + 3 F e^{3 +}$

2. In a second step, the total chromium content was measured by dissolving a crystal in $2.9 M H C L O_{4}$ at and cooling to $20 ° C$ . Excess and were then added to oxidize all chromium to $C r_{2} O_{7}^{2 -}$ . Unreacted $S_{2} {O_{8}}^{- 2}$ was destroyed by boiling, and the remaining solution was titrated with standard $F e^{2 +}$ . In this step, each Crin the original unknown reacts with three $F e^{2 +}$ .

$C r^{x +} + \overset{S_{2} O_{8}^{2 -}}{\to} C r_{2} O_{7}^{2 -} \frac{1}{2} C r_{2} O_{7}^{2 -} + 3 F e^{2 +} \to C r^{3 +} + 3 F e^{3 +}$

In Step 1,0.4375g of laser crystal required 0.498mL of (prepared by dissolving in ). In step , of crystal required of the same solution. Find the average oxidation number of in the crystal and find the total micrograms ofpre gram of crystal.

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.

Why don't ${Cr}^{3 +}$ and ${TiO}^{2 +}$ interfere in the analysis of ${Fe}^{3 +}$ when a Walden reductor, instead of a Jones reductor, is used for pre-reduction?

Would indigo tetrasulfonate be a suitable redox indicator for the titration of $Fe {(CN)}_{6}^{4 -}$ with ${TI}^{3 +}$ in 1MHCI? (Hint: The potential at the equivalence point must be between the potentials for each redox couple.)

See all solutions

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