The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation
Given that it requires 32.9 mL of 0.330 M Na2S2O3(aq) to titrate a 30.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.
mols S2O3^2- = M x L = ?
mols I3^- = 1/2 that (look at the ionic equation)
M = mols I3^-/L I3^-
Well, it seems like we have a titration situation here. Let's break it down!
According to the net ionic equation, the reaction between I3– and S2O32– can be represented as:
I3–(aq) + 2S2O32–(aq) → 3I–(aq) + S4O62–(aq)
From the stoichiometry of the reaction, we can see that it takes 2 moles of S2O32– to react with 1 mole of I3–. So, the moles of I3– in the 30.0 mL sample can be calculated as follows:
moles of I3– = (moles of S2O32–) / 2
To find the moles of S2O32–, we can use the following equation:
moles of S2O32– = (concentration of S2O32–) x (volume of S2O32–)
Now, let's substitute the given values:
moles of S2O32– = (0.330 M) x (32.9 mL) = 10.827 moles
Finally, we can determine the moles of I3–:
moles of I3– = (10.827 moles) / 2 = 5.414 moles
To find the molarity of I3– in the solution, we divide the moles of I3– by the volume of the sample in liters:
molarity of I3– = (moles of I3–) / (volume of sample in liters)
volume of sample = 30.0 mL = 0.030 L
Let's calculate the molarity:
molarity of I3– = (5.414 moles) / (0.030 L) ≈ 180.47 M
So, the molarity of I3– in the solution is approximately 180.47 M. And remember, if you meet any other titrations along the way, feel free to call me back, and I'll be here to sprinkle some amusing chemistry tricks on your questions!
To determine the molarity of I3–(aq), we can use the stoichiometry of the reaction and the given information about the titration.
The balanced net ionic equation for the reaction between I3–(aq) and S2O32–(aq) is:
2 S2O32–(aq) + I3–(aq) → 3 I–(aq) + S4O62–(aq)
From the equation, we can see that 2 moles of S2O32–(aq) react with 1 mole of I3–(aq). Therefore, the moles of I3–(aq) in the solution can be determined using the stoichiometry of the reaction.
Given:
Volume of Na2S2O3(aq) = 32.9 mL = 0.0329 L
Molarity of Na2S2O3(aq) = 0.330 M
Volume of I3–(aq) = 30.0 mL = 0.0300 L
Let's calculate the moles of S2O32–(aq) using the given information:
Moles of Na2S2O3 = Molarity × Volume
Moles of Na2S2O3 = 0.330 M × 0.0329 L
Moles of Na2S2O3 = 0.0108 mol
Since the stoichiometry of the reaction is 2:1 (S2O32–:I3–), the moles of I3– can be calculated as follows:
Moles of I3– = (0.0108 mol of Na2S2O3) ÷ 2
Moles of I3– = 0.0054 mol
Finally, we can calculate the molarity of I3–(aq):
Molarity of I3– = Moles ÷ Volume
Molarity of I3– = 0.0054 mol ÷ 0.0300 L
Molarity of I3– = 0.18 M
Therefore, the molarity of I3–(aq) in the solution is 0.18 M.
To calculate the molarity of I3–(aq) in the solution, we can use the information provided about the titration of the solution with Na2S2O3(aq).
First, let's write the balanced net ionic equation for the reaction between I3–(aq) and S2O32–(aq):
2S2O32–(aq) + I3–(aq) → S4O62–(aq) + 3I–(aq)
From the balanced equation, we can see that the stoichiometric ratio between S2O32– and I3– is 2:1. This means that it takes 2 moles of S2O32– to react with 1 mole of I3–.
Given that it requires 32.9 mL of 0.330 M Na2S2O3(aq) to titrate a 30.0 mL sample of I3–(aq), we can use this information to find the number of moles of S2O32– that reacted.
Number of moles of Na2S2O3(aq) = Molarity × Volume (in liters)
= 0.330 M × (32.9 mL / 1000 mL) [converting mL to L]
= 0.010857 moles
Since the stoichiometric ratio between S2O32– and I3– is 2:1, the number of moles of I3– is half of the number of moles of S2O32–.
Number of moles of I3–(aq) = 0.010857 moles / 2
= 0.0054285 moles
Now, we can calculate the molarity of I3–(aq) in the solution using the volume of the sample.
Molarity of I3–(aq) = Number of moles / Volume (in liters)
= 0.0054285 moles / (30.0 mL / 1000 mL) [converting mL to L]
= 0.1809 M
Therefore, the molarity of I3–(aq) in the solution is 0.1809 M.