Suppose a student diluted and titrated a bleach unknown exactly as described in the experimental procedure, except only a single titration was performed which required 13.73 mL of 0.100 M Na2S2O3.
The density of the original, undiluted bleach unknown was 1.04 g/mL
1. Calculate the number of moles of Na2S2O3 used in the titration
2. Calculate the number of moles of ClO- in the sample titrated. Hint: The moles of thiosulfate simply equals the moles of sodium thiosulfate. From the moles of thiosulfate the moles of I2 can be easily determined from reaction (2). Now use reaction (1) to get the moles of hypochlorite, ClO−.
3. Calculate the grams of NaClO in the titrated bleach sample, assuming all of the hypochlorite ion comes from sodium hypochlorite. Hint: The moles of sodium hypochlorite is simply the same as the moles of hypochlorite.
4. Calculate the mass in grams of UNDILUTED bleach that was in the sample of bleach titrated.
5. Using the UNROUNDED answers to questions 3 and 4, calculate the weight percent of NaClO in the undiluted bleach sample. Hint: Weight percent is the same thing as percent by mass.
1. mols = M Na2S2O3 x L Na2S2O3
2. What are you calling reaction 1 and reaction 2?
3,4,5. What is the volume of the bleach. How much was it diluted?
1. To calculate the number of moles of Na2S2O3 used in the titration, we use the formula:
moles of Na2S2O3 = volume of Na2S2O3 (L) x concentration of Na2S2O3 (M)
Given that 13.73 mL (or 0.01373 L) of 0.100 M Na2S2O3 was used, we can substitute these values into the formula:
moles of Na2S2O3 = 0.01373 L x 0.100 M = 0.001373 moles
Therefore, the number of moles of Na2S2O3 used in the titration is 0.001373 moles.
2. From the balanced equation of reaction (2), we know that:
Na2S2O3 + I2 -> Na2S4O6 + 2NaI
Since the moles of thiosulfate (Na2S2O3) used in the titration equals the moles of I2, we can use this information to determine the moles of I2. In this case, the moles of I2 is also the moles of ClO-.
Therefore, the number of moles of ClO- in the sample titrated is 0.001373 moles.
3. From the balanced equation of reaction (1), we know that:
2ClO- + 2NaI -> I2 + 2NaCl
Since the moles of ClO- (hypochlorite) react in a 1:1 ratio with moles of I2, the moles of ClO- is equal to the moles of NaI. Therefore, the moles of sodium hypochlorite (NaClO) is also equal to the moles of ClO-.
Hence, the number of moles of NaClO in the titrated bleach sample is 0.001373 moles.
4. To calculate the mass in grams of undiluted bleach, we use the formula:
mass of bleach (g) = volume of bleach (mL) x density of bleach (g/mL)
Since we know the density of the undiluted bleach is 1.04 g/mL, we need to convert the volume used in the titration (13.73 mL) to the volume of undiluted bleach.
Since the bleach was diluted, we need to calculate the dilution factor first. The dilution factor is the ratio of the volume of undiluted bleach to the volume used in the titration.
dilution factor = volume of undiluted bleach / volume used in titration
Given that the dilution factor is 20 (provided in the experimental procedure), we can calculate the volume of undiluted bleach as follows:
volume of undiluted bleach = dilution factor x volume used in titration = 20 x 13.73 mL = 274.6 mL
Substituting the volume (274.6 mL) and density (1.04 g/mL) into the mass calculation formula:
mass of undiluted bleach = 274.6 mL x 1.04 g/mL = 285.184 g
Therefore, the mass of undiluted bleach that was in the sample of bleach titrated is 285.184 grams.
5. To calculate the weight percent (percent by mass) of NaClO in the undiluted bleach sample, we divide the mass of NaClO in the undiluted bleach by the mass of undiluted bleach, and then multiply by 100.
mass of NaClO = mole of NaClO x molar mass of NaClO
Given that the molar mass of NaClO is 74.44 g/mol, we can calculate the mass of NaClO as follows:
mass of NaClO = 0.001373 moles x 74.44 g/mol = 0.1019 g
Substituting the values into the weight percent formula:
weight percent of NaClO = (0.1019 g / 285.184 g) x 100 = 0.0357 x 100 = 3.57%
Therefore, the weight percent of NaClO in the undiluted bleach sample is 3.57%.
To answer these questions, we need to follow a step-by-step approach. Let's break down each question and explain how to solve it.
1. Calculate the number of moles of Na2S2O3 used in the titration:
To calculate the number of moles of Na2S2O3 used, we need to multiply the volume of the solution (13.73 mL) by its concentration (0.100 M). This formula can be used:
Number of moles = Volume (in L) x Concentration (in mol/L).
First, convert the volume to liters by dividing it by 1000 (since 1 mL = 0.001 L). Then, multiply the volume in liters by the concentration to get the number of moles of Na2S2O3 used.
2. Calculate the number of moles of ClO- in the sample titrated:
Since the moles of thiosulfate (Na2S2O3) equals the moles of sodium thiosulfate, we can use this value to determine the moles of I2 (from reaction 2). Then, using reaction 1, we can find the moles of hypochlorite (ClO-). Use the balanced chemical equations to convert between moles of reactants and products.
3. Calculate the grams of NaClO in the titrated bleach sample:
To calculate the grams of NaClO, we need to know the molar ratio between NaClO and Na2S2O3. From the balanced chemical equation, we can determine this ratio. Multiply the moles of NaClO by its molar mass to get the mass in grams.
4. Calculate the mass in grams of UNDILUTED bleach:
To calculate the mass of undiluted bleach, we need to know the volume of the diluted solution and its density. We can use the equation: mass = volume x density.
5. Calculate the weight percent of NaClO in the undiluted bleach sample:
To find the weight percent, we need the mass of NaClO (from question 3) and the mass of the undiluted bleach (from question 4). Divide the mass of NaClO by the mass of undiluted bleach, and multiply by 100 to get the percentage.
Remember to use the given values and perform calculations step-by-step to find the answers to each question.