a student titrates three samples of 0.3% hydrogen peroxide solution made from a bottle of 3% hydrogen peroxide solution dated 10/31/2007. would you expect the calculated molarities of the samples to be higher or lower than the following values:
trial 1: .0015 molar
trial 2: .00135 molar
trial 3: .001425 molar
Well, let me check my crystal ball... Oh wait, I left it at the circus! Well, based on what you've given me, I would say that the calculated molarities of the samples are likely to be lower. After all, those samples have been sitting around since 2007! Time does pass, and hydrogen peroxide might not be as perky as it used to be. But don't lose hope, those values might still be good enough to put on a good fireworks show!
To determine whether the calculated molarities of the samples would be higher or lower than the provided values, we need to consider the dilution factor caused by adding water to the 3% hydrogen peroxide solution.
First, let's calculate the dilution factor:
Dilution factor = (Final volume)/(Initial volume)
For all three trials, the initial volume of the 3% hydrogen peroxide solution is the same, but the final volume is different based on the amount of water added.
Given that the samples are made from a 0.3% hydrogen peroxide solution, we can calculate the dilution factor:
Dilution factor = 0.3% / 3% = 0.1
This means that the final volume is 10 times the initial volume.
Now, let's calculate the molarities of the samples:
For trial 1:
Calculated molarity = (0.0015 M) x (10) = 0.015 M
For trial 2:
Calculated molarity = (0.00135 M) x (10) = 0.0135 M
For trial 3:
Calculated molarity = (0.001425 M) x (10) = 0.01425 M
Comparing these calculated molarities with the provided values:
- Trial 1: Expected higher than 0.0015 M
- Trial 2: Expected lower than 0.00135 M
- Trial 3: Expected lower than 0.001425 M
Therefore, we would expect the calculated molarities of the samples to be higher than the provided value for trial 1 and lower for trials 2 and 3.
To determine whether the calculated molarities of the samples would be higher or lower than the given values, we need to consider the concept of concentration and the effect of time on the stability of hydrogen peroxide.
Hydrogen peroxide can decompose over time, especially when exposed to light, heat, or certain catalysts. This decomposition reaction converts hydrogen peroxide into water and oxygen gas.
In this case, the bottle of 3% hydrogen peroxide solution is dated 10/31/2007, which indicates that it is more than ten years old. Considering the age of the solution, it is reasonable to expect that some decomposition may have occurred, leading to a decrease in the concentration of hydrogen peroxide.
Since the hydrogen peroxide concentration in the original solution decreases due to decomposition, the calculated molarities of the samples would be expected to be lower than the given values. This is because a lower concentration of hydrogen peroxide would result in a smaller number of moles in a given volume, leading to lower molarity values.
Therefore, the calculated molarities of the samples are expected to be lower than the following values:
Trial 1: 0.0015 molar
Trial 2: 0.00135 molar
Trial 3: 0.001425 molar
I found a site on the Internet that says "If you carefully read the label and the MSDS you will see that 3% H2O is 3% by weight". Well, I have a bottle of H2O2 at home and it doesn't say anything about what 3% is. Neither does the MSDS sheet on the web. I looked at every source I could find that sells 3% H2O2 and one of them said they sold 30%, 50% and 70% by weight That still doesn't tell me what the 3% is but if we go with what this guy says about "reading carefully" (his source listed 30% H2O2 as a density of 1.11 g/mL) which makes me think your post of a day or so that "assumes" 1.00 g/mL as density for 3% as reasonable.
So if 3% is 3% w/w that means 3g H2O2/100 g solution. And since it is diluted by a factor of 10 I think the 1.00 g/mL must be very close.
3g H2O2 is 3/34.01 = 0.0882 mols.
Mass of that solution will be 100 g and that will be volume of 0.1L.
M then is 0.0882/0.1 = 0.882
In the 0.3% solution, M = 0.0882 since it has been diluted by a factor of 10.
According to two sources on the net the decomposition of commercial H2O2 is very slow even over long periods of time due to stabilizers that are added to the solution. You may have some information in your course work that gives an indication of how much might be lost versus time; my best educated guess, if these net sources are correct is that the titrated samples would be higher than those listed in the problem. However, you may have some information that will contradict that. Feel free to disagree. I don't have any personal experience.