Lab equipment and a scary reality....

[BeanAnimal]

As we have established, chemistry makes my brain hurt.

Please correct me if I am wrong, as I am, not sure I am headed down the right path.

So with HCL, (M) Molarity and (N) Normality are the same?
So I can use simple math for dilution?
C1V1=C2V2

If I start with 1 M HCL and want 1000 mL of HCL at 0.1 M
1.0 M × V1 = 0.1 M × 1000 mL

Solve for V1
V1 = 0.1 M × 1000 mL / 1.0 M

SO
V1 = 100 mL of 1 M HCL

Equipment Needed
1000 mL Class A volumetric flask
100 mL Class A volumetric flask (or volumetric pipette)

Procedure
1 - Fill 100 mL flask to line (bottom of meniscus) with 1M HCL
2 - Add roughly 750 mL of lab grade distilled water to 1000 mL flask
3 - Slowly add the measured 100 mL of 1M HCL to the partially full 1000 mL flask
4 - Swirl to mix (or stopper and invert?)
5 - Fill 1000 mL flask to line.
6 - Stopper and invert 25 times (is that overkill?)

Accuracy
Assuming the beginning solution is actually 1 M and the water is lab grade, and both flasks are "Class A" and within their stated accuracy.
(Let's ignore me being able to hit the line precisely)

I read the tolerances of class A to be
100 mL Flask Tolerance: ±0.08 mL
1000 mL Flask Tolerance: ±0.30 mL

For the 100 ml flask that means an actual volume (excluding my precision) of
V1min = 100 mL − 0.08 mL = 99.92 mL
V1max = 100 mL + 0.08 mL = 100.08 mL

For the 1000 ml flask that means an actual final volume of (excluding my precision) of
V2min = 1000 mL − 0.30 mL = 999.7 mL
V2max = 1000 mL + 0.30 mL = 1000.3 mL

Calculation of final possible concentrations of my dilution using the equipment above and ignoring any precision error introduced by me:
Cfmin = 1.0 M × 99.92 mL / 1000.3 mL = 0.09992 M
Cfmax = 1.0 M × 100.08 mL / 999.7 mL = 0.10008 M

So a final acid concentration somewhere between 0.09992 M and 0.10008 M ??

In the real world (if I even remotely did that right) what would the final range (±) look like with the human element of precision added in reading the flask and/or other issues (a drop or two left in the 100 ml etc. ??

Sorry if this is basic or silly. In the end I want to know if reducing 1 M HCL to 0.1 M (N?) for use as a KH reagent (titration) is reasonable or will introduce too much error to be worth the trouble.

 

[taricha]

If you are diluting with lab volumetric flasks and good scales, then the errors involved in making stocks and doing dilutions etc is practically always much less than the error in whatever chemistry process you are using it for.
If that makes sense.

 

[BeanAnimal]

It makes sense. I guess then if the "Amazon" sold "Class A" equipment actually meets the spec silkscreened onto the bottle.

Are my calculations and assumptions correct?

I have not had good luck with the auto pipettes for sure.

 

[Randy Holmes-Farley]

The procedure is correct, and so are the calculations.

The uncertainly in concentration also needs to take the initial concentration uncertainty into account.

Unless you buy a very expensive standard, it’s uncertainty will be more than your calculations suggest from dilution errors.

 

[Dan_P]

As we have established, chemistry makes my brain hurt.

Please correct me if I am wrong, as I am, not sure I am headed down the right path.

So with HCL, (M) Molarity and (N) Normality are the same?
So I can use simple math for dilution?
C1V1=C2V2

If I start with 1 M HCL and want 1000 mL of HCL at 0.1 M
1.0 M × V1 = 0.1 M × 1000 mL

Solve for V1
V1 = 0.1 M × 1000 mL / 1.0 M

SO
V1 = 100 mL of 1 M HCL

Equipment Needed
1000 mL Class A volumetric flask
100 mL Class A volumetric flask (or volumetric pipette)

Procedure
1 - Fill 100 mL flask to line (bottom of meniscus) with 1M HCL
2 - Add roughly 750 mL of lab grade distilled water to 1000 mL flask
3 - Slowly add the measured 100 mL of 1M HCL to the partially full 1000 mL flask
4 - Swirl to mix (or stopper and invert?)
5 - Fill 1000 mL flask to line.
6 - Stopper and invert 25 times (is that overkill?)

Accuracy
Assuming the beginning solution is actually 1 M and the water is lab grade, and both flasks are "Class A" and within their stated accuracy.
(Let's ignore me being able to hit the line precisely)

I read the tolerances of class A to be
100 mL Flask Tolerance: ±0.08 mL
1000 mL Flask Tolerance: ±0.30 mL

For the 100 ml flask that means an actual volume (excluding my precision) of
V1min = 100 mL − 0.08 mL = 99.92 mL
V1max = 100 mL + 0.08 mL = 100.08 mL

For the 1000 ml flask that means an actual final volume of (excluding my precision) of
V2min = 1000 mL − 0.30 mL = 999.7 mL
V2max = 1000 mL + 0.30 mL = 1000.3 mL

Calculation of final possible concentrations of my dilution using the equipment above and ignoring any precision error introduced by me:
Cfmin = 1.0 M × 99.92 mL / 1000.3 mL = 0.09992 M
Cfmax = 1.0 M × 100.08 mL / 999.7 mL = 0.10008 M

So a final acid concentration somewhere between 0.09992 M and 0.10008 M ??

In the real world (if I even remotely did that right) what would the final range (±) look like with the human element of precision added in reading the flask and/or other issues (a drop or two left in the 100 ml etc. ??

Sorry if this is basic or silly. In the end I want to know if reducing 1 M HCL to 0.1 M (N?) for use as a KH reagent (titration) is reasonable or will introduce too much error to be worth the trouble.

I buy 0.1 N HCl but diluting 1 M to 0.1 M as you describe will work.

After you transfer the 100 mL of 1 M HCl solution to the 1 L flask, rinse the 100 mL flask several times with the water you are using for dilution to ensure that you are transferring every little bit of HCl to the 1 L flask.

Edit: Here is a link that explains how to calculate the propagation of error (not saying what you did is right or wrong)

Use your tolerances for standard deviation in this equation for the estimate.

​

Propagation of Error

Propagation of Error (or Propagation of Uncertainty) is defined as the effects on a function by a variable's uncertainty. It is a calculus derived statistical calculation designed to combine …

chem.libretexts.org

 

[BeanAnimal]

I buy 0.1 N HCl but diluting 1 M to 0.1 M as you describe will work.

I was doing that as well, but 0.1 N is $20 per liter and 1.0 N is 30 per liter... or $40 for 2 liters.

I use a bit less than liter per month at 4 test per day. So figure 10 liters per year. $200 vs $30 plus the cost of some mid range flasks and bottles.

I just don't want to go through the trouble and find out that I wasted my time. Nothing worse than aiming for some precision mark, only to be let down by improper tools or technique, be it wood working or volumetric mixing.

The plan is to purchase 2 liters of 1.0 n

Mix (10) 1 liter batches at a time and store them in the amber bottles.
I will transfer the spare 1 n from the plastic bottle to the glass Boston bottle as well for more stable storage, as I understand plastic is not good for long term storage of acid (gas permeation?).

Edit: Here is a link that explains how to calculate the propagation of error (not saying what you did is right or wrong)

Use your tolerances for standard deviation in this equation for the estimate.

​

Propagation of Error

Propagation of Error (or Propagation of Uncertainty) is defined as the effects on a function by a variable's uncertainty. It is a calculus derived statistical calculation designed to combine …

chem.libretexts.org

Thank you!

 

[BeanAnimal]

The procedure is correct, and so are the calculations.

The uncertainly in concentration also needs to take the initial concentration uncertainty into account.

Unless you buy a very expensive standard, it’s uncertainty will be more than your calculations suggest from dilution errors.

Thank you for the insight. In your opinion am I better of just buying the 0.1 N and dealing with the inherent error without adding additional error by dilution?

Would it be prudent to do some type of titration test to verify the 1 n if I were to use it?

 

[BeanAnimal]

BTW the title of this thread is wrong. I was going to opine about the poor result I have had with some Amazon import lab equipment in the past and the though that not only are many startups or poor labs likely using this stuff instead of Fisher, Corning, or whatever... but who also knows what other "major" suppliers are peddling it to to larger enterprise and labs.

 

[Randy Holmes-Farley]

Thank you for the insight. In your opinion am I better of just buying the 0.1 N and dealing with the inherent error without adding additional error by dilution?

Would it be prudent to do some type of titration test to verify the 1 n if I were to use it?

The 1.0 N is fine, but if the uncertainty of 0.1 N ranges from 0.95 to 1.05 N, that error is way bigger than your dilution.

One can also buy 1.00 or 1.000 N, but price goes up and may be harder to find.

 

[BeanAnimal]

So we infer the uncertainty (given a reputable vendor) by the number of decimal places listed or are you just giving an hypothetical (again sorry, I have no experience in any of this).

For example this form Hach product listed here is 1.00

Hydrochloric Acid Standard Solution, 1.00 N, 1 L

Hydrochloric Acid Standard Solution, 1.00 N, 1 L

www.hach.com

Am I to assume = 1.00 could be 1.005 to 0.995 ?

 

[Randy Holmes-Farley]

So we infer the uncertainty (given a reputable vendor) by the number of decimal places listed or are you just giving an hypothetical (again sorry, I have no experience in any of this).

For example this form Hach product listed here is 1.00

Hydrochloric Acid Standard Solution, 1.00 N, 1 L

Hydrochloric Acid Standard Solution, 1.00 N, 1 L

www.hach.com

Am I to assume = 1.00 could be 1.005 to 0.995 ?

It should be.

 

[Dburr1014]

As we have established, chemistry makes my brain hurt.

Please correct me if I am wrong, as I am, not sure I am headed down the right path.

So with HCL, (M) Molarity and (N) Normality are the same?
So I can use simple math for dilution?
C1V1=C2V2

If I start with 1 M HCL and want 1000 mL of HCL at 0.1 M
1.0 M × V1 = 0.1 M × 1000 mL

Solve for V1
V1 = 0.1 M × 1000 mL / 1.0 M

SO
V1 = 100 mL of 1 M HCL

Equipment Needed
1000 mL Class A volumetric flask
100 mL Class A volumetric flask (or volumetric pipette)

Procedure
1 - Fill 100 mL flask to line (bottom of meniscus) with 1M HCL
2 - Add roughly 750 mL of lab grade distilled water to 1000 mL flask
3 - Slowly add the measured 100 mL of 1M HCL to the partially full 1000 mL flask
4 - Swirl to mix (or stopper and invert?)
5 - Fill 1000 mL flask to line.
6 - Stopper and invert 25 times (is that overkill?)

Accuracy
Assuming the beginning solution is actually 1 M and the water is lab grade, and both flasks are "Class A" and within their stated accuracy.
(Let's ignore me being able to hit the line precisely)

I read the tolerances of class A to be
100 mL Flask Tolerance: ±0.08 mL
1000 mL Flask Tolerance: ±0.30 mL

For the 100 ml flask that means an actual volume (excluding my precision) of
V1min = 100 mL − 0.08 mL = 99.92 mL
V1max = 100 mL + 0.08 mL = 100.08 mL

For the 1000 ml flask that means an actual final volume of (excluding my precision) of
V2min = 1000 mL − 0.30 mL = 999.7 mL
V2max = 1000 mL + 0.30 mL = 1000.3 mL

Calculation of final possible concentrations of my dilution using the equipment above and ignoring any precision error introduced by me:
Cfmin = 1.0 M × 99.92 mL / 1000.3 mL = 0.09992 M
Cfmax = 1.0 M × 100.08 mL / 999.7 mL = 0.10008 M

So a final acid concentration somewhere between 0.09992 M and 0.10008 M ??

In the real world (if I even remotely did that right) what would the final range (±) look like with the human element of precision added in reading the flask and/or other issues (a drop or two left in the 100 ml etc. ??

Sorry if this is basic or silly. In the end I want to know if reducing 1 M HCL to 0.1 M (N?) for use as a KH reagent (titration) is reasonable or will introduce too much error to be worth the trouble.

This makes your head hurt??

I couldn't even get past the first 5 sentances and got completly lost.

LOL