select from the following the most appropriate concentration for NaOH to titrate a sample that contains 250mg of aspirin:
a. 0.01 M
b. 0.10 M
or
c. 1.0 M
and explain why you chose it.
Well, that's a good question! Now, let's break it down.
To determine the appropriate concentration of NaOH for titrating a sample containing 250mg of aspirin, we need to consider a few factors. The most important thing to keep in mind is that we want an accurate and efficient titration process, without using an excessive amount of the reagent.
If we opt for a very low concentration, like 0.01 M (option a), then we might have to use a larger volume of NaOH to neutralize the sample. This could lead to longer titration times and potentially inaccurate results. So, I'm going to rule out option a.
Now, let's look at option b: 0.10 M NaOH. This concentration is a bit higher, which means we would need a smaller volume of NaOH to react with the sample. This could speed up the titration process and potentially provide more accurate results. It seems like a reasonable choice, but let's continue evaluating our options.
Lastly, option c suggests a 1.0 M NaOH solution. This concentration is quite high, and although it would allow for rapid titration, it may also increase the risk of overshooting the end point and introducing errors into the analysis. Plus, it's always wise to work with lower concentrations when possible to minimize potential safety hazards.
Taking all of this into consideration, I would choose option b, 0.10 M NaOH. It strikes a balance by providing a reasonably fast and accurate titration without the need for excessive amounts of NaOH.
Remember, titration can be quite a balancing act, just like juggling chainsaws while riding a unicycle! I hope this answer didn't clown around too much with your understanding.
To determine the most appropriate concentration of NaOH to titrate a sample containing 250 mg of aspirin, we need to consider the molar amount of aspirin and the stoichiometry of the reaction.
The molecular weight of aspirin (C9H8O4) can be calculated as follows:
C: 12.01 g/mol x 9 = 108.09 g/mol
H: 1.01 g/mol x 8 = 8.08 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total molecular weight = 108.09 g/mol + 8.08 g/mol + 64.00 g/mol = 180.17 g/mol
To convert the mass of aspirin (in mg) to moles, we divide by the molar mass:
250 mg / 180.17 g/mol = 1.387 x 10^-3 mol
The balanced equation for the reaction between NaOH and aspirin is:
C9H8O4 + NaOH -> C9H7O4Na + H2O
From the balanced equation, we can see that it takes one mole of NaOH to react with one mole of aspirin.
Since the molar amount of aspirin in the 250 mg sample is 1.387 x 10^-3 mol, we need an equal or slightly greater molar amount of NaOH to ensure complete reaction.
Considering the concentrations given, let's calculate the molar amount of NaOH in each solution:
a. 0.01 M NaOH:
1 mole/L x 0.01 L = 0.01 mol
b. 0.10 M NaOH:
1 mole/L x 0.10 L = 0.10 mol
c. 1.0 M NaOH:
1 mole/L x 1.0 L = 1.0 mol
Based on the molar amounts calculated, the most appropriate concentration of NaOH would be option c. 1.0 M. This concentration provides a sufficient excess of NaOH to react completely with the 1.387 x 10^-3 mol of aspirin present in the sample.
To determine the most appropriate concentration for NaOH to titrate a sample containing 250 mg of aspirin, we need to consider the balanced chemical equation for the titration reaction between NaOH and aspirin.
The balanced equation for the reaction between NaOH and aspirin (acetylsalicylic acid) is as follows:
C9H8O4 (aspirin) + NaOH → C9H7O4Na (sodium acetylsalicylate) + H2O
From the equation, we can see that the mole ratio between NaOH and aspirin is 1:1. This means that for every mole of aspirin, we need one mole of NaOH to react completely.
To calculate the molar mass of aspirin:
C9H8O4 -> (9 * 12.01 g/mol) + (8 * 1.008 g/mol) + (4 * 16.00 g/mol) = 180.16 g/mol
From the molar mass, we can determine the number of moles of aspirin in the sample:
Number of moles = Mass / Molar mass = 250 mg / 180.16 g/mol = 0.00138 mol
Since the molar ratio between NaOH and aspirin is 1:1, we need an equivalent number of moles of NaOH to achieve complete titration. Therefore, the most appropriate concentration of NaOH can be determined by considering the desired volume of the NaOH solution.
Suppose we choose a 50 mL volume of NaOH solution for titration. To determine the needed concentration, divide the number of moles of NaOH required by the volume in liters:
Concentration (M) = Number of moles / Volume (L) = 0.00138 mol / 0.050 L = 0.0276 M
Among the given options:
a. 0.01 M: This concentration would provide insufficient NaOH for complete titration, as we require 0.0276 M NaOH. Hence, this option is not appropriate.
b. 0.10 M: This concentration would provide more than enough NaOH for complete titration. While we could use this concentration, it would be excessive and potentially lead to overshooting the endpoint during titration.
c. 1.0 M: This concentration is significantly higher than the required concentration. It is not necessary to use such a concentrated solution for this titration. Additionally, using a highly concentrated NaOH solution can be hazardous.
Considering these factors, the most appropriate concentration for NaOH to titrate a sample containing 250 mg of aspirin is option b, 0.10 M. However, it is important to note that adjusting the volume of the NaOH solution may be necessary to achieve the desired stoichiometric ratio.
250 mg aspirin = 0.250/molar mass.
mols aspirin = mols NaOH required.
Therefore, M = mols NaOH/L NaOH and I can't answer this question without knowing the size buret you will use.
In fact I would not want to use any of them.
I was always taught that if we used a 50 mL buret that we want to use about 40 mL to titrate a sample.
0.00139/0.04L = about 0.035M and that isn't one of the answers. Here is what I would do.
Calculate the volume each of the answers would give you and make a choice from that.
M = mols/L or L = mols/M
The first one gives 0.00139/0.01 = 0.139 or 139 mL and that means more than one 50 mL buret of NaOH. You never want to use more than one buret of stuff.
Next is 0.00139/0.1 = 0.0139 or about 14 mL. Frankly I wouldn't choose this one because this increases the error when small volumes are used; however, this probably is the best choice of the three.
The final one is 0.00139/1 = 0.00139 L or 1.39 mL which is FAR FAR too small. .I suspect you are to choose the middle one.
You might ask what I would do. I would use about 3x the sample of 750 mg which gives 0.750/180 = about 0.0042 mol and
0.0042 mols/0.01M = 0.042 L or about 42 mL.