Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?

Explanation:

The given chemical reaction is:

From above equation   serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.

Given that :

the volume of potassium iodide

the molarity of potassium

the volume of distilled water

The volume of 3%

Molarity of AgNO₃

Let take an integral look with the reaction between KI and AgNO₃; we have

At the end point; the moles of KI will definitely be equal to the moles of AgNO₃

So;

Thus; the volume of 0.1 M AgNO₃  needed to reach the end point is 15 mL

to make a supersaturated solution, make a saturated solution of sugar by adding 360 grams of sugar to 100 ml of water at 80 degrees celsius. when the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water should only dissolve 210 grams of sugar.