A 25 ml sample of household bleach was diluted to 500 ml in a volumetric flask. An unmeasured excess of potassium iodide was aded to a 20 ml aliquot of the diluted sample; the iodine liberated in the reaction: OCl- + 2I- -> I2 + Cl- + H2O
required 34.5 ml of 0.0409 M Na2S2O3. Calculate the weight- volume precentage of NaOCl in the sample.
i just get stuck in the middle of the problem :S
The titration is: 2 Na2S2O3 + I2 = Na2S4O6 + 2 NaI
moles NaOCl in 20mL aloquat
= moles I2
= (1/2)moles Na2S2O3
= 0.345 L * 0.0409 M / 2
moles NaOCl in original (25mL) sample
= 0.0345 * 0.0409 * (500/20) mol / 2
mass NaOCl in original (25mL) sample
= 74.4424 g/mol * 0.0345 L * 0.0409 M * 12.5
%w/v of NaOCl in original sample
= gram nass solute per 100mL solution
= 74.4424 * 0.0345 * 0.0409 * 12.5 * 4 g/100mL
= 5.25... g/100mL
= 5.25... %w/v
Thank you so so so much! You're great!
To calculate the weight-volume percentage of NaOCl in the sample, we need to determine the amount of NaOCl in the 25 ml sample.
Step 1: Determine the amount of Na2S2O3 used in the reaction.
The balanced equation for the reaction shows that each mole of iodine, I2, reacts with 2 moles of Na2S2O3. Therefore, the mole ratio between I2 and Na2S2O3 is 1:2.
Given that 0.0409 M Na2S2O3 was used, and the volume used was 34.5 ml, we can calculate the number of moles of Na2S2O3 used.
Moles of Na2S2O3 = Molarity x Volume
Moles of Na2S2O3 = 0.0409 mol/L x 0.0345 L
Moles of Na2S2O3 = 0.00141 moles
Since the mole ratio between I2 and Na2S2O3 is 1:2, this means that 0.00141 moles of I2 were produced in the reaction.
Step 2: Determine the amount of NaOCl in the 25 ml sample.
Since the 25 ml sample was diluted to 500 ml, the concentration of NaOCl in the diluted sample is 25/500 times the concentration in the original sample.
Let's say the concentration of NaOCl in the 25 ml sample is C mol/L. Then,
C x 25 ml = concentration of NaOCl in the diluted sample x 500 ml
We can set up the equation as follows:
C x 25 = concentration of NaOCl in the diluted sample x 500
We know that 0.00141 moles of I2 reacted with the Na2S2O3 used in the titration. According to the balanced equation, this means that 0.00141 moles of NaOCl were present in the 25 ml sample.
Therefore,
C x 25 ml = 0.00141 moles
C = 0.00141 moles / 25 ml
Step 3: Calculate the weight of NaOCl in the 25 ml sample.
The molar mass of NaOCl is 74.44 g/mol.
Weight of NaOCl = Number of moles x Molar mass
Weight of NaOCl = 0.00141 moles x 74.44 g/mol
Now, you can calculate the weight-volume percentage of NaOCl in the sample using the following formula:
Weight-volume percentage (%) = (Weight of NaOCl / Volume of sample) x 100
Remember to convert the volume of the sample to milliliters (ml) if needed.
To solve this problem, we need to follow a step-by-step approach. Let's break it down into smaller parts.
Step 1: Calculate the moles of Na2S2O3 used.
We are given that 34.5 ml of 0.0409 M Na2S2O3 is used. The volume of the Na2S2O3 solution is given in milliliters (ml), so we need to convert it to liters (L).
34.5 ml = 34.5/1000 L = 0.0345 L
Now we can calculate the moles of Na2S2O3 used:
moles of Na2S2O3 = volume (L) × concentration (M)
moles of Na2S2O3 = 0.0345 L × 0.0409 M
Step 2: Calculate the moles of I2.
We know from the balanced equation that the reaction between Na2S2O3 and I2 has a 1:2 stoichiometric ratio. This means that one mole of Na2S2O3 reacts with two moles of I2.
moles of I2 = 2 × moles of Na2S2O3
Step 3: Calculate the moles of NaOCl.
We know from the balanced equation that the reaction between NaOCl and I2 has a 1:1 stoichiometric ratio. This means that one mole of NaOCl reacts with one mole of I2.
moles of NaOCl = moles of I2
Step 4: Calculate the weight of NaOCl.
To calculate the weight of NaOCl, we need to know its molar mass. The molar mass of NaOCl is given by:
Na = 22.99 g/mol
O = 16.00 g/mol
Cl = 35.45 g/mol
molar mass of NaOCl = 22.99 + 16.00 + 35.45 g/mol
weight of NaOCl = moles of NaOCl × molar mass of NaOCl
Step 5: Calculate the weight-volume percentage of NaOCl.
The weight-volume percentage is calculated by dividing the weight of NaOCl by the volume of the sample and multiplying by 100.
weight-volume percentage of NaOCl = (weight of NaOCl / volume of sample) × 100
Now let's plug in the values and calculate the answer.