Sodium thiosulfate, Na2S2O3, is used as a fixer in photographic film developing. The amount of Na2S2O3 in a solution can be determined by a titration with iodine, I2, according to the equation: 2Na2S2O3(aq) + I2(aq) --> Na2S4O6 +2NaI(aq).

Calculate the concentration of the Na2S2O3 solution if 21.40 mL of a 0.1350 M I2 solution react exactly with a 100.0 mL sample of the Na2S2O3 solution. Use 4 sig. fig.

Aside, the end of the titration is determined by the color. NaI is a pale yellow and I2 is a deep purple. Just when the purple color persists, all the iodine that can react has reacted.

To calculate the concentration of the Na2S2O3 solution, we can use the stoichiometry of the balanced equation and the volume and concentration of the I2 solution.

Given:
Volume of I2 solution = 21.40 mL
Concentration of I2 solution = 0.1350 M
Volume of Na2S2O3 solution = 100.0 mL

Step 1: Determine the moles of I2 used in the reaction.
Moles of I2 = volume (in L) x concentration
= 21.40 mL x (1 L / 1000 mL) x 0.1350 M
= 0.0028905 moles

Step 2: Use the stoichiometry of the balanced equation to determine the moles of Na2S2O3 reacted.
From the balanced equation, we can see that 2 moles of Na2S2O3 react with 1 mole of I2.
Therefore, moles of Na2S2O3 = 0.0028905 moles x (2 moles Na2S2O3 / 1 mole I2)
= 0.005781 moles

Step 3: Calculate the concentration of the Na2S2O3 solution.
Concentration (in M) = moles / volume (in L)
= (0.005781 moles) / (100.0 mL x (1 L / 1000 mL))
= 0.05781 M

Therefore, the concentration of the Na2S2O3 solution is 0.05781 M, rounded to 4 significant figures.

To calculate the concentration of the Na2S2O3 solution, you can use the stoichiometry of the balanced equation and the volume and concentration of the I2 solution used.

First, let's write down the balanced equation for the reaction:
2Na2S2O3(aq) + I2(aq) → Na2S4O6(s) + 2NaI(aq)

From the equation, we can see that 2 moles of Na2S2O3 react with 1 mole of I2. Therefore, the number of moles of I2 used can be calculated using the following equation:

moles of I2 = concentration of I2 (M) × volume of I2 solution (L)

Substituting the given values:
moles of I2 = 0.1350 M × 0.02140 L = 0.002889 moles of I2

Since the stoichiometry of the reaction is 2:1 between Na2S2O3 and I2, the number of moles of Na2S2O3 is half of the moles of I2:

moles of Na2S2O3 = 0.002889 moles of I2 / 2 = 0.001445 moles of Na2S2O3

Next, we need to calculate the concentration of Na2S2O3. We have the number of moles of Na2S2O3 and the volume of the Na2S2O3 solution used.

Using the equation:

concentration of Na2S2O3 (M) = moles of Na2S2O3 / volume of Na2S2O3 solution (L)

Substituting the values:
concentration of Na2S2O3 = 0.001445 moles / 0.1000 L = 0.01445 M

Rounding to four significant figures:
concentration of Na2S2O3 = 0.01445 M

Therefore, the concentration of the Na2S2O3 solution is 0.01445 M.

2S2O3^2- + I2 ==> 2I^- + S4O6^2-

1. mols I2 = M x L = ?
2. Convert mols I2 to mols S2O3^2-. That is ?mols I2 x (2 mol S2O3^2-/1 mol I2) = ?
3. Then M S2O3^2- = mols S2O3^2-/L S2O3^2-