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Why double amount of potassium iodide is required for dissolving iodine?
It sounds like you're describing Lugol's iodine. It's not technically "required," it's just that's what the recipe is. Actually, in Lugol's iodine, KI is needed to improve the dissolution of iodine in water.
Why potassium iodate is added to potassium iodide in iodometry titration?
Iodometric titration means the titration of triiodide. Preparation of triiodide can be prepared by adding solid iodine to excess potassium iodide, however since iodine sublimes it is more difficult to weigh than potassium iodate. The triiodide would then need to be standardized with primary standard grade arsenious oxide. A better way is to weigh potassium iodate and add to a small excess of potassium iodide. The two ions (iodide and iodate) will also form the needed triiodide in an acidic environment. If the reagent is made fresh this way it can be used to standardize thiosulfate. Iodide iodate of known normality can also be used to titrate unknown concentrations of sulfite. For example, steam boiler treatment applications.
Why is a high temperature needed to melt potassium iodide?
Potassium iodide has a high melting point because of its strong ionic bonds between potassium and iodine atoms. To overcome these bonds and separate the ions, a high temperature is required to provide enough energy to break the bonds and transition the solid to a liquid state.
How many atoms of potassium are needed to react with an atom of iodine?
One atom of potassium will react with one atom of iodine according to the balanced chemical equation for the reaction between potassium and iodine: 2K + I2 → 2KI.
How many moles of potassium iodide are needed to make 750 ml of a 1.8 M solution?
To find the number of moles of potassium iodide needed, multiply the volume of the solution (750 ml) by the molarity (1.8 moles/L). First, convert the volume to liters (750 ml = 0.75 L), then multiply 0.75 L by 1.8 moles/L to get 1.35 moles of potassium iodide.
Why double amount of potassium iodide is required for dissolving iodine?
It sounds like you're describing Lugol's iodine. It's not technically "required," it's just that's what the recipe is. Actually, in Lugol's iodine, KI is needed to improve the dissolution of iodine in water.
Why do we rinse the leaf I water?
The leaf was rinsed in water to rehydrate it. Iodine solution is an aqueous solution of iodine/potassium iodine - potassium tri-iodide; water is needed inside the leaf to enable penetration by diffusion.
What is the reason behind using 40 g of potassium iodide and 12.7 g of iodine to prepare 0.1 N of iodine?
- The atomic weight of iodine is 126,90447; for a 0,1 N solution, dividing by 10 the result is 12,69. - The iodide (KI) is added to increase the solubility of iodine in water or alcohol.
Is potassium iodide ionic or molecular?
Potassium is K1+, and Iodine is I1-. As a result, Potassium iodide is made. It's ionic becuase Potassium has a positive charge, so it needs one more electron to have its orbitals filled. Iodine on the other hand, has an electron that isn't needed. If Iodine can give up its electron, then all of its orbitals will be filled. As a result, iodine gives it's extra electron to the potassium, and they both have filled orbitals. When an electron is being given and accepted, that's called an ionic bond. So Potassium iodide is ionic.
What does iodized salt contain?
Iodized salt is a combination of regular table salt, sodium chloride, with a small amount of a salt of iodine, potassium iodide. Iodized salt has gained popularity in some countries because iodine is needed by the thyroid to function properly and iodine deficiency can cause the thyroid disease called goiter.
Why potassium iodate is added to potassium iodide in iodometry titration?
Iodometric titration means the titration of triiodide. Preparation of triiodide can be prepared by adding solid iodine to excess potassium iodide, however since iodine sublimes it is more difficult to weigh than potassium iodate. The triiodide would then need to be standardized with primary standard grade arsenious oxide. A better way is to weigh potassium iodate and add to a small excess of potassium iodide. The two ions (iodide and iodate) will also form the needed triiodide in an acidic environment. If the reagent is made fresh this way it can be used to standardize thiosulfate. Iodide iodate of known normality can also be used to titrate unknown concentrations of sulfite. For example, steam boiler treatment applications.
How many grams of potassium iodide will produce 750 grams of silver iodide when there is an excess of silver nitrate?
530,3 g potassium iodide are needed.
Why is a high temperature needed to melt potassium iodide?
Potassium iodide has a high melting point because of its strong ionic bonds between potassium and iodine atoms. To overcome these bonds and separate the ions, a high temperature is required to provide enough energy to break the bonds and transition the solid to a liquid state.
Add an excess of potassium iodide to copper sulphate solution how much is excess?
The amount of excess potassium iodide depends on the stoichiometry of the reaction between potassium iodide and copper sulfate. One equivalent of potassium iodide is needed to react with one equivalent of copper sulfate. Excess potassium iodide would be any amount added beyond this stoichiometric ratio.
Why the melting point of potassium fluoride is higher than the melting point of potassium iodide?
KF has small size than KCl. So the packing of atoms / ions in KF is more than that in KCl. So a larger energy is needed to separate the atoms from solid state to liquid state and hence KF has higher melting point than KCl.
How many atoms of potassium are needed to react with an atom of iodine?
One atom of potassium will react with one atom of iodine according to the balanced chemical equation for the reaction between potassium and iodine: 2K + I2 → 2KI.
How many iodine ions are needed to balance out a barium ion?
A barium ion has a charge of 2+ and an iodide ion has a charge of 1-, therefore, two iodide ions are needed to balance the barium ion. Ba2+ + 2I- --> BaI2
