Hi, I'm really having trouble this this chemistry problem. I have the answers; however I don't have extensive knowledge about titrations and I'm struggling with working it out. If anyone could suggest how to work this out (or even where to start) that would be great. Thanks!
Triiodide ions are generated in a solution by the following reaction in acidic solution.
IO3- (aq) + I- ----> I3-
Triiodie ion is determined by titration with a thiosulfate solution (Na2S2O3). The products are iodide ion and tetrathionate ion (S4O6 2+).
a. Balance the equation for the reaction of IO3- with I- ions.
b. Write and balance the equation for the reaction of S2O3 -2 with I3- in acidic solution.
c. A 25.00 mL sample of a .01000 M solution of KIO3 is reacted with excess KI. It requires 32.04 mL of Na2S2O3 to titrate the I3- ions present. What is the molarity of the Na2S2O3 solution?
d. how would you prepare 500.0 mL of the KIO3 solution in part c, using pure dry solid KIo3.
Yes, I know you must use the half reaction method to balance the equation. I put a 2 as the coefficient of I- but I'm not sure how to divide the equation into two reactions since there are only 3 parts.
Let's do this bit by bit.
The IO3^-/I^- is an oxidation/reduction reaction. Do you know how to balance that type? The I in IO3^- has an oxidation state of +5 and the I^- is -1.
Personally, I think using I3^- complicates a relatively easy equation to write. So let's use I2 and we can add I^- to it later if that's what the professor wants.
6H^+ + IO3^- + 5I^- ==> 3I2 + 3H2O Then the liberated I2 is titrated with thiosulfate in this reaction.
I2 + 2S2O3^-2 ==> S4O6^-2 + 2I^-
This gets part a and b EXCEPT if the prof wants I3^-, then just add I^- to make it work. For example, the iodate equation, replace 3I2 with 3I3^- and balance that by using 8I^- on the left instead of 5I^-. For the second one use 3I^- on the right and I3^- on the left.
For part c. calculate mols IO3^-, use the equation to convert to mols I2 and use the titration equation to convert to mols thiosulfate. Then L x M = mols. You have mols thiosulfate and L thioulfatre and you can calculate M thiosulfate.
For part d. remember M = mols/L. You want 0.01 mols in 1 L so you will need -.005 mols in 500 mL. grams = mols x molar mass.
thank you so much!
wait, i'm sorry, do i get the moles of IO3 by divinding 174.9 (molar mass of IO3) by 1196.16g (the mass of the IO3, hydrogen and I3 combined?) (the whole left side of the equation?)
To work through this chemistry problem, let's break it down step by step:
a. Balancing the equation for the reaction of IO3- with I- ions:
The reaction is already given as: IO3- (aq) + I- ----> I3-
To balance the equation, first, we need to make sure the number of iodine atoms is the same on both sides. Here we have 1 iodine atom on both sides, so that's already balanced.
Now let's balance the oxygen atoms. On the left side, we have 3 oxygen atoms from IO3- and 0 oxygen atoms from I-. On the right side, we have 0 oxygen atoms from I3-. To balance it, we need to add 3 H2O molecules to the right side:
IO3- (aq) + I- + 3H2O ----> I3- + 3H2O
The equation is now balanced.
b. Writing and balancing the equation for the reaction of S2O3 -2 with I3- in acidic solution:
The reaction is given as:
I3- + S2O3 -2 ----> I- + S4O6 2-
To balance this equation, we need to make sure the number of iodine atoms and the number of sulfur atoms are equal on both sides.
On the left side, we have 1 iodine atom from I3- and 2 sulfur atoms from S2O3 -2. On the right side, we have 1 iodine atom from I- and 2 sulfur atoms from S4O6 2-. The equation is already balanced.
c. Determining the molarity of Na2S2O3 solution:
We are given that 32.04 mL of Na2S2O3 is required to titrate the I3- ions. We also know the initial concentration and volume of KIO3 solution.
First, calculate the number of moles of I3- ions present in the 25.00 mL of 0.01000 M KIO3 solution:
moles of I3- ions = concentration (M) × volume (L)
= 0.01000 M × 0.02500 L
= 0.000250 mol
Since 1 mol of I3- ions reacts with 2 mol of Na2S2O3 according to the balanced equation, we can set up a proportion to find the molarity of Na2S2O3:
(0.000250 mol I3-) / (2 mol Na2S2O3) = (x mol Na2S2O3) / (0.03204 L Na2S2O3)
Solving for x, we get:
x = (0.000250 mol I3-) × (2 mol Na2S2O3) / (0.03204 L Na2S2O3)
x ≈ 0.0155 M
Therefore, the molarity of the Na2S2O3 solution is approximately 0.0155 M.
d. Preparing 500.0 mL of the KIO3 solution:
To prepare a solution, there are a few important steps to follow:
1. Calculate the amount of KIO3 needed:
We are given the desired volume of the solution (500.0 mL) and the concentration of the solution (0.01000 M). Using the equation:
amount (mol) = concentration (M) × volume (L)
amount (mol) = 0.01000 M × 0.5000 L
amount (mol) = 0.00500 mol
2. Calculate the mass of KIO3 needed:
To calculate the mass of KIO3 needed, we use the molar mass of KIO3. The molar mass of KIO3 is the sum of the atomic masses of potassium (K), iodine (I), and oxygen (O) in KIO3.
3. Convert moles to grams:
mass (g) = amount (mol) × molar mass (g/mol)
4. Dissolve the calculated mass of KIO3 in water:
Add the calculated mass of KIO3 to a container and add water gradually while stirring until the desired volume of 500.0 mL is reached.
Note: Make sure to handle KIO3 carefully, as it is an oxidizing agent and can cause harm if mishandled.
I hope this explanation helps you understand the steps to solve these chemistry problems. If you have any further questions, feel free to ask!