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11 Dec 2019

Aqueous potassium iodate (KIO3) and potassium iodide (KI) react in the presence of dilute hydrochloric acid, as shown below.

KIO3(aq) + 5KI(aq) + 6HCl(aq) ? 3I2(aq) + 6KCl(aq) + 3H2O(l)

What mass of iodine (I2) is formed when 50.0 mL of 0.020 M KIO3 solution reacts with an excess of KI and HCl?

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Trinidad Tremblay
Trinidad TremblayLv2
13 Dec 2019

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Related questions


A saturated solution of calcium iodate, Ca(IO3)2, is created by suspending a small amount of pure solid of Ca(IO3)2 in 50.0 mL of distilled water. After allowing equilibrium to be established, a 10.0-mL sample of the saturated solution is pipeted into an Erlenmeyer flask, and about 20 mL of de-ionized water is added to this sample, followed by about 1 cm3 of solid KI and 1 mL of 2.0 M HCl. The addition of excess KI and HCl converts all of the IO3- present in the sample to molecular iodine according to the following equation.IO3-(aq) + 5 I-(aq) + 6H+(aq) ? 3 I2(aq) + 3H2O(l) (Eqn-1) The iodine produced by the above reaction is then titrated with 0.0250 M Na2S2O3 using starch indicator, which requires 29.0 mL of 0.0250 M Na2S2O3 to reach end-point. The stoichiometric reaction between iodine and thiosulfate is represented by the following net ionic equation.I2(aq) + 2 S2O32

I'm trying to find the amount of ethanol in an unkown solution by titration but I'm having trouble calculating it. Molarity of thiosulfate: 0.0806 M, amount of dichromate used: 15 mL

here is what my lab manual says:

The titration is based on the oxidation of ethanol with an oxidizing agent, such as potassium dichromate. Depending on the solution pH and the amount of potassium dichromate added, ethanol can be oxidized to ethanal (acetaldehyde, CH3CHO) via reaction 1, or to ethanoic acid (acetic acid, CH3COOH) via reaction 2:

3C2H5OH(aq) + Cr2O72-(aq) + 8H+ ? 3CH3CHO(aq) + 2Cr3+(aq) + 7H2O (1)

3C2H5OH(aq) + 2Cr2O72- (aq) + 16H+ ? 3CH3COOH(aq) + 4Cr3+ (aq) + 11H2O (2)

In order to ensure all ethanol has completely reacted via reaction 2, a known and excess amount of potassium dichromate (let’s call it potassium dichromate total) is added into a sample solution containing ethanol. The amount of unreacted potassium dichromate is then determined by reacting with potassium iodide solution via reaction 3 to form iodine:

Cr2O72- (aq) + 14H+ + 6I-(aq) ? 2Cr3+ (aq) + 7H2O(l) + 3I2 (aq) (3)

From the reaction 3, we know that 1 mole of dichromate ions can produce 3 moles of I2. Therefore, we just need to determine how much moles of I2 is produced in reaction 3, then, we can get the moles of unreacted potassium dichromate. Subtracting the moles of unreacted potassium dichromate from the moles of potassium dichromate total gives the moles of potassium dichromate that have reacted with ethanol. Then, from the stoichiometry of reaction 2, one should be able to know how much ethanol is in the test solution.

How to know the moles of I2 produced in reaction 3? The I2 is titrated with sodium thiosulfate (Na2S2O3) via the following reaction 4:

I2(aq) + 2S2O32-(aq) ? 2I-(aq) + S4O62-(aq) (4)

Experimental data: For the standardization of thiosulfate it took 31.7 mL of thiosulfate to reach the end point, for the unknown solution, it took 23.85 mL of thiosulfate to reach the endpoint. How do I use these numbers to find the amount of ethanol in the unknown solution?

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