How do I figure these out?
A solution of sodium thiosulfate, Na2S2O3, is 0.1365 M. 29.28 mL of this solution reacts with 39.79 mL of I2 solution. What is the molarity of the I2 solution?
2(Na2S2O3) + I2↔Na2S4O6 + 2(NaI)
16.77 mL of the I2 solution from above is required to titrate a sample containing As2O3. Calculate the mass of As2O3 (197.8 g/mol) in the sample.
As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI
To figure out the molarity of the I2 solution and the mass of As2O3 in the sample, you can use the concept of stoichiometry and the given information about the reaction equations and volumes.
1. Molarity of the I2 Solution:
First, let's use the given reaction equation and volumes to determine the molarity of the I2 solution.
From the equation: 2(Na2S2O3) + I2 ↔ Na2S4O6 + 2(NaI)
We can see that the ratio of Na2S2O3 to I2 is 2:1.
Given:
Molarity of Na2S2O3 solution = 0.1365 M
Volume of Na2S2O3 solution used = 29.28 mL
Volume of I2 solution used = 39.79 mL
Using the molarity and volume of Na2S2O3, we can calculate the number of moles of Na2S2O3 used:
moles of Na2S2O3 = molarity * volume
moles of Na2S2O3 = 0.1365 M * 0.02928 L (convert mL to L by dividing by 1000)
moles of Na2S2O3 = 0.00399792 mol
Since the ratio of Na2S2O3 to I2 is 2:1, the moles of I2 used is half of the moles of Na2S2O3 used:
moles of I2 = moles of Na2S2O3 / 2
moles of I2 = 0.00399792 mol / 2
moles of I2 = 0.00199896 mol
Now, we can calculate the molarity of the I2 solution using the moles of I2 and the volume used:
Molarity of I2 solution = moles of I2 / volume
Molarity of I2 solution = 0.00199896 mol / 0.03979 L (convert mL to L by dividing by 1000)
Molarity of I2 solution = 0.0502 M
Therefore, the molarity of the I2 solution is 0.0502 M.
2. Mass of As2O3 in the Sample:
Next, let's use the given reaction equation and volume of I2 solution to calculate the mass of As2O3 in the sample.
From the equation: As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI
We can see that the ratio of As2O3 to I2 is 1:2.
Given:
Volume of I2 solution used = 16.77 mL
Using the molarity and volume of I2, we can calculate the number of moles of I2 used:
moles of I2 = molarity * volume
moles of I2 = 0.0502 M * 0.01677 L (convert mL to L by dividing by 1000)
moles of I2 = 0.000840954 mol
Since the ratio of As2O3 to I2 is 1:2, the moles of As2O3 in the sample is half of the moles of I2 used:
moles of As2O3 = moles of I2 / 2
moles of As2O3 = 0.000840954 mol / 2
moles of As2O3 = 0.000420477 mol
Finally, we can calculate the mass of As2O3 using the moles of As2O3 and the molar mass:
Mass of As2O3 = moles of As2O3 * molar mass
Mass of As2O3 = 0.000420477 mol * 197.8 g/mol
Mass of As2O3 = 0.08323 g
Therefore, the mass of As2O3 in the sample is 0.08323 grams.
To figure out the molarity of the I2 solution and the mass of As2O3 in the sample, we can use the stoichiometry and the given information. Let's solve each problem step by step.
1. Calculating the molarity of the I2 solution:
The balanced equation for the reaction between Na2S2O3 and I2 is:
2(Na2S2O3) + I2 ↔ Na2S4O6 + 2(NaI)
From the balanced equation, we can see that the stoichiometric ratio between Na2S2O3 and I2 is 2:1. So, the number of moles of I2 can be calculated using the following formula:
moles of I2 = (moles of Na2S2O3) / 2
Given that the concentration of Na2S2O3 solution is 0.1365 M and the volume of Na2S2O3 solution is 29.28 mL (which we need to convert to liters), we can calculate the moles of Na2S2O3:
moles of Na2S2O3 = (concentration of Na2S2O3) * (volume of Na2S2O3 in liters)
= 0.1365 M * (29.28 mL / 1000 mL)
= 0.00399792 moles
Now, we can calculate the moles of I2:
moles of I2 = (moles of Na2S2O3) / 2
= 0.00399792 moles / 2
= 0.00199896 moles
Next, we need to find the volume of the I2 solution. The volume of I2 solution is 39.79 mL (which we need to convert to liters).
volume of I2 solution = 39.79 mL / 1000 mL
= 0.03979 L
Finally, we can find the molarity of the I2 solution using the following formula:
Molarity of I2 solution = (moles of I2) / (volume of I2 solution in liters)
= 0.00199896 moles / 0.03979 L
= 0.0502 M
Therefore, the molarity of the I2 solution is 0.0502 M.
2. Calculating the mass of As2O3 in the sample:
The balanced equation for the reaction between As2O3, I2, and H2O is:
As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI
From the balanced equation, we can see that the stoichiometric ratio between As2O3 and I2 is 1:2. We also know that 16.77 mL of the I2 solution was required to titrate the sample.
Now, let's find the moles of I2 using the molarity of the I2 solution we calculated earlier:
moles of I2 = (molarity of I2 solution) * (volume of I2 solution in liters)
= 0.0502 M * (16.77 mL / 1000 mL)
= 0.000841554 moles
Since the stoichiometric ratio between As2O3 and I2 is 1:2, the moles of As2O3 will be half of the moles of I2:
moles of As2O3 = (moles of I2) / 2
= 0.000841554 moles / 2
= 0.000420777 moles
Finally, we can calculate the mass of As2O3 using the molar mass of As2O3 (197.8 g/mol):
mass of As2O3 = (moles of As2O3) * (molar mass of As2O3)
= 0.000420777 moles * 197.8 g/mol
= 0.08324496 g
Therefore, the mass of As2O3 in the sample is approximately 0.0832 g.