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This is a titration problem.
Q. What is the molarity of 4% w/v acetic acid (molar mass = 60.05g/mol)
So, does 4% w/v acetic acid mean 4g of acetic acid per 100mL of vinegar?
This is my working out:
n=m/M
=4/60.05
=0.066mol
n=CV
C=n/V
= 0.066/0.1L
= 0.66M
Is that right?
Thank-you!
That's the definition of "w/v": mass per 100 mL of solution. Remember that since water weights about 1 gram/mL, 100 mL of any relatively dilute solution will weigh approximately 100 g. So, someone way back got real lazy and... that's where they came up with it.Quote:
Originally Posted by RosSin
n=m/M
=4/60.05
=0.066mol
That's correct. That's the number of moles per 100 mL (since you have 4 grams per 100 mL).
n=CV
C=n/V
= 0.066/0.1L
= 0.66M
Looks good to me.
Ok. Thanks for that.
The second part to this question is as follows:
Q. Calculate the volume of 0.10M NaOH you would need to react with 20.00mL of this vinegar.
My working out is as follows.
I've written out the equation: NaOH + CH3COOH ----> CH3COONa + H2O
which can be simplified as: CH3COOH + OH- ------> CH3COO- + H2O
So first I worked out the number of mole of vinegar: n = 0.66M (from previous question) * 0.02L = 0.013mol
As ratio of CH3COOH:NaOH is 1:1, n(NaOH) = 0.013mol
n = CV
V (NaOH) = 0.013/0.1 = 0.133L = 133.2mL NaOH
I just wanted to ask if everything is right. Thanks!
The third part to this question asks:
Q. The answer from (b). Is too large to be delivered from a burette accuretely; and would require a large amount of NaOH to be used in the lab. Therefore, we are going to dilute the vinegar so that we will use only around 20mL of NaOH from the burette. (The optimum burette titre is 15-20mL). Calculate by what amount you need to dilute the vinegar so that its final concentration is around 0.10M.
I'm not sure how to answer this question but I do know that I have to use the dilution factor.
Is the dilution factor C1V1 = C2V2?
Thanks!
The rest also appears to be correct.
The dilution factor is pretty simple. You can simply think like this:
Yes, C1V1 = C2V2 is the equation that you can use. You can also think like this: "I have a 0.66 M solution. I need a 0.1 M solution." The dilution factor is, therefor, 0.66 / 0.1 = 6.6. So, dilute it 1-to-6.6. (If you had a 660 mL flask, you would add 100 mL to the flask and then dilute it with another 560 mL to the flask. 660 mL volumetric flasks aren't generally available.).
Note that this is pretty hard to do. In the "real world", we usually don't use a burette to deliver to a volumetric flask. We would take an available pipette and an available volumetric flask; pipette from the original solution into the vol flask and dilute to the mark. Pipettes are typically available in 1, 5, 10, 50, 100, 500, etc mL. Volumetric flasks are available in 10, 50, 100, 500, 1000, etc. mL so you have to pick what's available.
Ok.
So, by using the dilution factor, my working out is:
C1V1=C2V2
0.66M * 0.1L = 0.1M * V2
V2 = 666mL
So, does that mean that 566mL of water has to be poured into a 100mL solution of vinegar to dilute it.
Thanks!
Yes -- or you pour 100 mL into a flask and add water so the final volume is 666 (that's the usual way it's done with volumetric flasks).
hi have similar question to the titration on the website.
calculate molarity of 4% w/v acetic acid ( Mr Ch3COOH=60.05g g/mol
You should post new questions as new threads.Quote:
Originally Posted by ene23
The problem you asked is the problem that we just solved in this thread.
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