TrueTears question thread

[TrueTears]

A sample of the energy drink High Caff was analysed for the concentration of caffeine it contained. Details on the pack indicated that it should contain 12 mg caffeine per 100 mL. A chemist decided to analyse a sample of the drink, without dilution, by HPLC. 100 mL of 10.0 stock standard solution of caffeine was prepared from pure caffeine tablets. The chemist decided to prepare standards of the following concentration: 5, 10, and 20 mg/100 mL caffeine. Available were 10 mL and 20 mL pipettes and 100 mL and 200 mL volumetric flasks.

a) Describe the dilutions the chemist would need to carry out in order to prepare the standard solutions using the equipments available.
b) The chemist could have used a 1 mL pipette and 100 mL volumetric flask to obtain a 10mg/100mL standard from the stock solution with only one dilution. Why is this a less accurate method?

[TrueTears]

hey guys, the questions i've posted since the holidays have all been answered, so need to look at them anymore XD

[TrueTears]

The concentration of a 2M sulfuric acid could also be correctly expressed as:
a 1.96 g/100mL
b 1960 ppm
c 19.6 % w/v
d 19.6 g/L

i picked c, is that right?

[TrueTears]

In an attempt to find the concentration of ethanoic ( acetic ) acid in vinegar, a student titrated a 20.00mL sample of a standard sodium hydroxide solution with a diluted vinegar solution from a burette. The student obtained the following titration results: 18.65 mL, 19.15 mL, 19.20 mL and 19.10 mL.

The discrepancy in the first titration could be due to the student washing the:
a) conical flask with the sodium hydroxide solution only.
b) pipette with water only
c) burette with water only
d) pipette with sodium hydroxide solution only
e) burette with diluted vingar solution only

umm.. for this question is the diluted vinegar solution in a conical flask or in the burette? I thought normally the unknown concentration sample would be in the conical and the standard solution would be in the burette, but the question says "...diluted vinegar solution from a burette", and what would the answer be?

[TrueTears]

A manufacturer claims that a sample of lead nitrate, , is 95.7% pure. A student wishes to check the validity of this claim by carrying out a gravimetric analysis. He aims to precipitate the lead ions as lead iodide ( has a solubility of 0.76 ).

The student carefully and accurately weights 15.07 g of the sample and dissolves it in water in a clean volumetric flask. He makes the volume up to 250.0 mL exactly. Using a clean, dry pipette he measures 20.00mL samples of this solution into three clean flasks. To each flask he adds an excess of sodium iodide solution. Lead iodide precipitates.

For each experiment he collects all of the solid and washes it carefully and thoroughly with water. Each sample is then dried for 1 hr at 90 deg C. Lead iodide does not decompose at this temperature.

1.Determine the mass of lead iodide the student might expect to obtain from each of the experiments.

2. The student obtained the following results:
mass of collected: 1.524g, 1.743g, 1.458g.
Suggest two reasons to explain the variation in his results.

Any help is appreciated !!!!

[jackinthepatch]

I got 0.8390g of PbI₂
So 0.84g correct to significant figures.

Is that right?

mass of collected: 1.524g, 1.743g, 1.458g.

My value is less than what the question is asking...I must have made an error somewhere lol...

[TrueTears]

soz, there are no answers to these questions, thats why i want to see if anyone can get similar answers etc. haha

[jackinthepatch]

What did you get for the mass of PbI₂ TrueTears?

[TrueTears]

lol mine is definitely wrong, i got 0.100 g, actually how would you do this question?

Basically i've wrote the equation and tried some stuff... but i keep getting answers that way off the results that the student got. lol

[jackinthepatch]

Haha. Yeah we're both way off what the student got. Um...well I wrote the equation of the lead nitrate reacting with the sodium iodide, then used stoich to work out the mass of the lead iodide. Oh, of course you had to make sure you divided the original mass of 15.07g by 12.5, because the student only used the 20mL of solution, not the 250mL that contained the whole 15.07g. The whole thing was pretty easy to calculate...maybe I'm missing a step...mmm. Lol.

[TrueTears]

Haha. Yeah we're both way off what the student got. Um...well I wrote the equation of the lead nitrate reacting with the sodium iodide, then used stoich to work out the mass of the lead iodide. Oh, of course you had to make sure you divided the original mass of 15.07g by 12.5, because the student only used the 20mL of solution, not the 250mL that contained the whole 15.07g. The whole thing was pretty easy to calculate...maybe I'm missing a step...mmm. Lol.

yeah i did all that but doesn't it say the sample is 95.7% pure? so you would use 95.7% OF 15.07 g...? or is that wrong?

[jackinthepatch]

Oh yeah good point lol, I totally missed that. Also I'm not sure whether or not the 0.76g/L solubility is needed for anything either haha.

Maybe Mao can help us, I hear he's alright at chem. Lol.

[TrueTears]

Oh yeah good point lol, I totally missed that. Also I'm not sure whether or not the 0.76g/L solubility is needed for anything either haha.

Maybe Mao can help us, I hear he's alright at chem. Lol.

hahah yeah i tried to figure around with that 0.76 g/L as well, didn't get anywhere LOL

[TrueTears]

Also can someone check if this question i did is right

An approximately 0.3M sodium hydroxide (NaOH) solution was standardised by titrating it against a standard 0.123 M sulfuric acid () solution. A 20.00mL aliquot of sodium hydroxide solution required an average titre of 23.65 mL.



Calculate the molarity of the NaOH solution.

My solution is as follows:

n() = 0.123 x 0.02365 = 0.0030745 mol
therefore n(NaOH) = 2 x 0.0030745 = 0.006149 mol
[NaOH] = = 0.30745 M = 0.307 M (3 s.f)

Also what does "An approximately 0.3M sodium hydroxide (NaOH) solution was standardised..." mean?

Thanks guys!!

[jackinthepatch]

I did this:

C_NaOH x V_NaOH= 2 x C_H2SO4 x V_H2SO4


C_NaOH = 2 x 0.123 x 0.02365 / .02
                   
Which then gives .291M.

I did all the calculations in one step though...which may explain the difference between our two answers. Maybe? Lol.