Acid is Hydroiodic Acid (HI) and the Base (Alkali) is potassium hydroxide.
Here is the reaction eq'n.
HI(aq) + KOH(aq) = KI(aq) + H2O(l).
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
The product of aqueous chlorine reacting with aqueous potassium iodide is potassium chloride and iodine. The chlorine oxidizes the iodide ions to form iodine, while the potassium ions from potassium iodide combine with the chlorine ions to form potassium chloride.
Yes, the amount of potassium iodide added in potassium (V) iodate solution will affect the amount of iodine liberated because potassium iodide reacts with potassium (V) iodate to produce iodine. Increasing the amount of potassium iodide will result in more iodine being liberated.
a precipitate. motha nacha -.0
To prepare a 5% potassium iodide solution, weigh 5 grams of potassium iodide and dissolve it in 100 mL of water. Stir until the potassium iodide is completely dissolved to achieve a 5% solution.
It is the Iodine dissolved in aqueous Potassium(or Sodium) Iodide
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
The product of aqueous chlorine reacting with aqueous potassium iodide is potassium chloride and iodine. The chlorine oxidizes the iodide ions to form iodine, while the potassium ions from potassium iodide combine with the chlorine ions to form potassium chloride.
When ferric chloride (FeCl3) is added to a solution of potassium iodide (KI), it reacts to form iron(III) iodide (FeI3) and potassium chloride (KCl). The iron(III) iodide produced is a brownish-red color, indicating the presence of the Fe3+ ion. This reaction can be represented by the following chemical equation: 2FeCl3 + 6KI -> 2FeI3 + 6KCl
Yes, the amount of potassium iodide added in potassium (V) iodate solution will affect the amount of iodine liberated because potassium iodide reacts with potassium (V) iodate to produce iodine. Increasing the amount of potassium iodide will result in more iodine being liberated.
Yes, it is correct.
a precipitate. motha nacha -.0
To prepare a 5% potassium iodide solution, weigh 5 grams of potassium iodide and dissolve it in 100 mL of water. Stir until the potassium iodide is completely dissolved to achieve a 5% solution.
If more potassium iodide is added to the potassium iodate (V) solution in the conical flask, there will be more iodine liberated. This is because potassium iodide reacts with potassium iodate (V) to produce iodine. Therefore, increasing the amount of potassium iodide increases the rate of reaction and the amount of iodine generated.
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.
Starch is composed of amylose and amylopectin, and is not soluble in water due to the presence of amylopectinIodine (I₂) is somewhat soluble in water, but is more soluble in iodide (I⁻) solutions, such as potassium iodide solution (KI).Aqueous iodine molecules (I₂) and iodide ions (I⁻) together will form triiodide ions (I₃⁻), which can react with amylose found in starch to produce a deep-blue colour in the solution. So all of iodide (I⁻), iodine (I₂) and amylose (or starch) are required together to produce the colour.This can be used to test for:Amylose/Starch: Add iodine dissolved in potassium iodide solution to test solution, orIodine: Add starch and potassium iodide solutions to test solution.If the substance being tested for is present, then triiodide ions (I₃⁻) can react with amylose (in starch) to produce a deep-blue colour, that is, a positive result.