[Yuti]

So this was the experiment I had to do, I just want to see if someone can check if my response is correct

Material 2 measuring cups 100 ml - 2 Erlenmeyer flasks 100 ml - Balance - Filter paper - Pasteur pipette - Pipette balloon - Funnel - Filling pipette 20 ml - Filling pipette 10 ml Fabrics FFT (0.05 %) – ready to use copper sulfate pentahydrate solution (0.5 M) -

Calculate how many grams of copper sulfate pentahydrate you need to weigh in a 200 mL volumetric flask to make a 0.5 mol/L solution

I already did this.

Write down these calculations below. Liquid drain unblocker – diluted 10 times - Write down how you can make this. I already did this

The final volume of the diluted solution is 100 mL. Safety - waste management - safety glasses - Find out for yourself which P and H sentences belong to the products and where the waste streams belong. Mention this in your report! 5. Method and observations

Ø Carefully pour +/- 50 mL of diluted drain cleaner into the measuring cup. Pipette 20 ml unblocker from this into the Erlenmeyer flask. Add 2 drops of FFT. - Describe what you see in the photo.

Ø Add a few drops of CuSO4 solution. - Describe what you see in the photo Ø Continue to add CuSO4 solution until the color of the precipitate changes from dark blue to light blue.

Ø Filter the solution and collect the filtrate in a new Erlenmeyer flask. Rinse the first Erlenmeyer flask with distilled water and filter this as well. Let the filter paper with the residue dry for several days. Determine as accurately as possible the mass of the filter paper with the residue. Record the mass. m (residue) = 0.5g

Ø Pipette 10 mL of undiluted drain cleaner into a measuring cup. Record the mass of 10 mL of drain cleaner. m (10 mL drain cleaner) = 9.9 g (to be filled in yourself)

6. Calculations Ø Calculate stoichiometrically the concentration of NaOH in the liquid drain cleaner, expressed in M. Take into account that the liquid drain cleaner was first diluted. Use the following response: 2 NaOH + CuSO4 à Cu(OH)2 + Na2SO4

7. Decision Ø Formulate an appropriate decision

8. Reflection Ø Calculate the molar concentration of NaOH in the undiluted drain cleaner. Ø Calculate the density of the undiluted drain cleaner. Ø Calculate the mass percentage of sodium hydroxide in the undiluted drain cleaner





My answer:


Cu(OH)2 -------- so I did:
cn=m/M.V----- 0.5g/97,561.0,020l=0.2562499359

Since the reaction is 2 NaOH + CuSO4 à Cu(OH)2 + Na2SO4
I think it's just: 0.2562499359 X2=0.5124998718 X10 (I diluted it 10 times but I want to under-thin)=5.12499871 mol/l=NaOH
Neat is: 5.124998718 mol/l NaOH
The density of undiluted: m/V ------ 9.9g/0.1L=99g/L
Mass percentage of sodium hydroxide in the undiluted drain cleaner:
0.1 l mass undiluted = 9.9 g
mass of NaOH in 0.1l undiluted drain cleaner:
cn=n/V------cn.V=n
5.124998718 mol/l . 0.1l = 0.512499871 mol
n=m/M --------n.M=m
m=0.512499871 mol. 39.997g/mole=20.499994872g

But than when I do 20,4g/9,9 X100=207 percent, but why? Where is my mistake?

[mjc123]

9.9 g is the mass of 10 mL. Is that 0.1 L?

[Yuti]

9.9 g is the mass of 10 mL. Is that 0.1 L?

It is the mass of 10ml undiluted water. You're right ! It is 0,01L and 20ml.is.0,02 L! Stupid mistake...Can I post my new calculations and would you like to check them please?

[mjc123]

It all looks pretty correct to me apart from that mistake, where you get a factor of 10 out at the end. 10 mL contains about 2 g NaOH.
(Oh, and you rounded 20.4999... to 20.4 g - careful!)

[Yuti]

Calculate the molar concentration of NaOH in the undiluted drain cleaner.
Cu(OH)2
Given: m=0.50g M=97.6 g/mol V=0.02l
Gevr: cn of NaOH=?
Sol: cn= m/M.V—-----------------0.50g/97.6g/mol.0.02l=0.26 mol/l
NaOH=0.26 mol/l X2= 0.52 mol/l X10=5.2 mol/l

Calculate the density of the undiluted drain cleaner.
Given: m=9.9g V=0.01l
Fr: ρ=?
 ρ=m/V—---------------------- 9.9g/0.01l=990g/l
Calculate the mass percentage of sodium hydroxide in the undiluted drain cleaner.
Geg: cn NaOH= 5.12 mol/l ρ drain cleaner =990g/L
Fr: m% NaOH
Sol: NaOH= 5.12 mol/l X 40g/mol= 204.8 g/l
m%= 204.8g/990g X100=20.7%

Is it now good?

[Yuti]

It all looks pretty correct to me apart from that mistake, where you get a factor of 10 out at the end. 10 mL contains about 2 g NaOH.
(Oh, and you rounded 20.4999... to 20.4 g - careful!)

"where you get a factor of 10 out at the end", I did that because they asked for undulited? Shouldn't I multiply by 10?

[mjc123]

No, you were a factor of 10 out because you said 10 mL was 0.1 L, not 0.01 L. That's all.
Now your calculations look right. i haven't checked them in detail, but they look in the right ball park.