13 replies to this topic

[fernandoz]

Hi there!,

Well, I've been doing some calculations to know the amount of Soda Ash and Sodium Hydroxide needed to raise the pH level of 6,000 gallons of waste water.

I did a kind of jar test to calculate these amounts, but I just want to check if I'm doing it in the right way (I saw some info. in the internet, and I think I'm doing something wrong).


Well, the initial pH level of the sample of 100 mL of waste water was 4.63

I started using a solution of .1 M of NaOH ( I added 4 gr of NaOH in 1 L of pure water, that's .1 M right?)

Then, after adding slowly the solution of NaOH into the waste water, it turned that I needed around 30 mL of a solution of .1 M NaOH to raise the pH level from 4.63 to 6.50 of a sample of 100mL of waste water

So here is the thing: if I want to know how much kg of NaOH should I add to adjust the pH level of 6,000 gallons, is the next calculation correct?

if:
1 L –----- 4 gr
.3 L ------ x

(if in one liter of pure water you have 4 grams of NaOH, then how many grams of NaOH do you have in a 30 mL sample?)

x = (.30 L * 4 gr) / 1 L
x = .12 gr of NaOH needed for 100 mL of waste water

and then if:

.12 gr ----- . 1 L (100 mL of waste water)
x ----- 6,000 gallons (22,712.4 L)

x = (22,712. 4 L ) * (.12 gr) / .1 L

x = 27,254.88 gr = 27.25 kg of NaOH for 6,000 gallons.

Since I was not that sure about my calculation, I did some research on the internet: and I found a different way to calculate it, there was a similar example but using other formulas: N1V1 = N2V2,
1 liter of disolution (50%) = 19.13 m NaOH = 19.13 N @ 1.53 gr of density

(.1)(w) = (19.13)(x)
x = (.1)*(w) / 19.13

x = volume of NaOH @ 50% in mL
w = volume of NaOH 0.1 N in mL

100 mL ------ x
y ----- z

z = (y * x) /100 mL

z = (0.1 * y * w) / 1913

z = liters of NaOH solution @ 50% needed to adjust the pH level of 6,000 gallons of waste water
y = total volume of waste water = 6,000 gallons
w = mL of .1 N NaOH needed to adjust the pH level of 100 mL of waste water = 30 mL

and it resulted an amount of 35.61 liters of NaOH (50% w/w) ... that's like around 17 kg of NaOH. (50% w/w means 50 gr of NaOH in 100 mL of water)

So, what am I doing wrong, what is the difference??

What about for Soda Ash,
I did the same procedure.
I measured 10.59 gr of Soda Ash in 1 Liter of water to have a .1 M solution.
And it resulted that I need around 20 mL of the solution to raise the pH from 4.63 to 6.50

Doing it my way:

1 L – 10.59 gr
.25 L - x

x = (.25 L * 10.59 gr) / 1L
x = .264 gr of soda ash for 100 mL of waste water

so:
.264 gr ----- 100 mL (.1 L)
x ----- 6,000 gallons (22,712.4 L)

x = (22,712. 4 L ) * (.264 gr) / .1 L

x = 59,960.736 gr = 59.96 kg of Soda Ash for 6,000 gallons?????

any idea of how to do it with the n1v1=n2v2 formula?
Sorry, for this long postt..
thankss!!

[breizh]

Hi there!,

Well, I've been doing some calculations to know the amount of Soda Ash and Sodium Hydroxide needed to raise the pH level of 6,000 gallons of waste water.

I did a kind of jar test to calculate these amounts, but I just want to check if I'm doing it in the right way (I saw some info. in the internet, and I think I'm doing something wrong).


Well, the initial pH level of the sample of 100 mL of waste water was 4.63

I started using a solution of .1 M of NaOH ( I added 4 gr of NaOH in 1 L of pure water, that's .1 M right?)

Then, after adding slowly the solution of NaOH into the waste water, it turned that I needed around 30 mL of a solution of .1 M NaOH to raise the pH level from 4.63 to 6.50 of a sample of 100mL of waste water

So here is the thing: if I want to know how much kg of NaOH should I add to adjust the pH level of 6,000 gallons, is the next calculation correct?

if:
1 L –----- 4 gr
.3 L ------ x

(if in one liter of pure water you have 4 grams of NaOH, then how many grams of NaOH do you have in a 30 mL sample?)

x = (.30 L * 4 gr) / 1 L
x = .12 gr of NaOH needed for 100 mL of waste water

and then if:

.12 gr ----- . 1 L (100 mL of waste water)
x ----- 6,000 gallons (22,712.4 L)

x = (22,712. 4 L ) * (.12 gr) / .1 L

x = 27,254.88 gr = 27.25 kg of NaOH for 6,000 gallons.

Since I was not that sure about my calculation, I did some research on the internet: and I found a different way to calculate it, there was a similar example but using other formulas: N1V1 = N2V2,
1 liter of disolution (50%) = 19.13 m NaOH = 19.13 N @ 1.53 gr of density

(.1)(w) = (19.13)(x)
x = (.1)*(w) / 19.13

x = volume of NaOH @ 50% in mL
w = volume of NaOH 0.1 N in mL

100 mL ------ x
y ----- z

z = (y * x) /100 mL

z = (0.1 * y * w) / 1913

z = liters of NaOH solution @ 50% needed to adjust the pH level of 6,000 gallons of waste water
y = total volume of waste water = 6,000 gallons
w = mL of .1 N NaOH needed to adjust the pH level of 100 mL of waste water = 30 mL

and it resulted an amount of 35.61 liters of NaOH (50% w/w) ... that's like around 17 kg of NaOH. (50% w/w means 50 gr of NaOH in 100 mL of water)

So, what am I doing wrong, what is the difference??

What about for Soda Ash,
I did the same procedure.
I measured 10.59 gr of Soda Ash in 1 Liter of water to have a .1 M solution.
And it resulted that I need around 20 mL of the solution to raise the pH from 4.63 to 6.50

Doing it my way:

1 L – 10.59 gr
.25 L - x

x = (.25 L * 10.59 gr) / 1L
x = .264 gr of soda ash for 100 mL of waste water

so:
.264 gr ----- 100 mL (.1 L)
x ----- 6,000 gallons (22,712.4 L)

x = (22,712. 4 L ) * (.264 gr) / .1 L

x = 59,960.736 gr = 59.96 kg of Soda Ash for 6,000 gallons?????

any idea of how to do it with the n1v1=n2v2 formula?
Sorry, for this long postt..
thankss!!

Hi Fernandoz ,
What you have to do is to take a sample of the WW you want to adjust the ph ( let say 500 ml) and introduce the reagents Na2CO3 or NaOH to reach the Ph ( from initial value 4.63 to final value Ph6 ) . Draw the titration curve ( ph versus quantity of reagent added ) . Finally you will have to do the mass balance for the 6000 gallons WW based on the sample you have tested (500 ml). It works .

Regards
Breizh

[fernandoz]

Hi Breizh!
Thanks for your reply.

So, by WW you mean to weigh the sample of 500mL?
For example, in the experiment I did, I had a sample of 100 mL of waste water.

I reached the desirable pH adding 20 ml of Soda Ash (.1 M Na2CO3)
If I use NaOH, then I'll be needing 30 mL .1 NaOH.

Thanks,
could you give me any clue about which formula should I use??


Hi Fernandoz ,
What you have to do is to take a sample of the WW you want to adjust the ph ( let say 500 ml) and introduce the reagents Na2CO3 or NaOH to reach the Ph ( from initial value 4.63 to final value Ph6 ) . Draw the titration curve ( ph versus quantity of reagent added ) . Finally you will have to do the mass balance for the 6000 gallons WW based on the sample you have tested (500 ml). It works .

Regards
Breizh
[/quote]

[breizh]

Fernandoz :
Caustic soda is very nasty (harmful) . Protective equipment is a must . You will need around 27 kg for neutralisation of 6000 gallons.

if Soda ash , let you try light soda ash cheaper than dense soda ash . Quantity needed around 48 kg for 6000 galons.

Regards
Breizh

[fernandoz]

Hi Breizh

Thanks a lot for your info...
I will take that on consideration (cautics is ver exothermic).

Could you give me a clue, how did you get those amounts? 27kg and 58 kg??
I would like to know how was the calculation, so in future I will not make the same mistakes.

thanks again

[Dacs]

I started using a solution of .1 M of NaOH ( I added 4 gr of NaOH in 1 L of pure water, that's .1 M right?)

My apologies if I haven't read your thread thoroughly, but I think the definition of Molarity is mol solute / L solution

I think the proper way of creating a 0.1 M NaOH (aq) is placing 1 mole (just calculate the equivalent mass) of NaOH (s) and diluting it to create 1 L of NaOH (aq). Solids have volume too and I think you need to consider this as well.

Your previous method will effectively produce a molarity lower than expected.

[Patrician]

Greetings!
That's correct fernandoz you used 0,1M (if you added 100% pure NaOH )
Since I'm a CHE student I think I'm closest to concentration matters.
Simple way to calculate concentration

c=n/V where n-number of mole, V-volume of solvent

n=m/M ,M for NaOH is 40g/mole

so for 0,1M=(4g/40g)/1L

including volume of NaOH is not permitted

I reached the desirable pH adding 20 ml of Soda Ash (.1 M Na2CO3)
If I use NaOH, then I'll be needing 30 mL .1 NaOH.

One thing you must check is purity of your soda ( NaOH is more basic than Na2CO3).
It's a common mistake to treat 70% NaOH as 100% pure one.

[breizh]

fernandoz :

6000 US gallons = 22.7 m3

Let 's talk Na2CO3 :
MW = 106g/mol
you used 20 ml of 0.1 mol/l to neutralize 100 ml of WW

0.1 mol/l = 10.6 g/l Na2CO3
20 ml of solution 0.1 m/l is equal to 20*10.6/1000 = 0.212 g of Na2CO3 used to neutralise 100 ml of WW

Now for 6000 US G= 0.212 * 22700/0.1 = 48124 g or 48.124 kg

regards
Breizh

[Dacs]

Greetings!
That's correct fernandoz you used 0,1M (if you added 100% pure NaOH )
Since I'm a CHE student I think I'm closest to concentration matters.
Simple way to calculate concentration

c=n/V where n-number of mole, V-volume of solvent

n=m/M ,M for NaOH is 40g/mole

so for 0,1M=(4g/40g)/1L

including volume of NaOH is not permitted

I reached the desirable pH adding 20 ml of Soda Ash (.1 M Na2CO3)
If I use NaOH, then I'll be needing 30 mL .1 NaOH.

One thing you must check is purity of your soda ( NaOH is more basic than Na2CO3).
It's a common mistake to treat 70% NaOH as 100% pure one.

I think that you being a student just reinforces the fact that you need to review your basics.

Let me repeat, unless that I'm grossly wrong with what I know about concentration, molarity is defined as moles of solute / L of solution.

By solution, it means that you must account for the volume of both solute and solvent.

If this would matter on dilute solutions (~0.1 M), I cannot say, but nevertheless, the volume of solute in a solution is not that negligible.

[Patrician]

Greetings!
That's correct fernandoz you used 0,1M (if you added 100% pure NaOH )
Since I'm a CHE student I think I'm closest to concentration matters.
Simple way to calculate concentration

c=n/V where n-number of mole, V-volume of solvent

n=m/M ,M for NaOH is 40g/mole

so for 0,1M=(4g/40g)/1L

including volume of NaOH is not permitted

I reached the desirable pH adding 20 ml of Soda Ash (.1 M Na2CO3)
If I use NaOH, then I'll be needing 30 mL .1 NaOH.

One thing you must check is purity of your soda ( NaOH is more basic than Na2CO3).
It's a common mistake to treat 70% NaOH as 100% pure one.

I think that you being a student just reinforces the fact that you need to review your basics.

Let me repeat, unless that I'm grossly wrong with what I know about concentration, molarity is defined as moles of solute / L of solution.

By solution, it means that you must account for the volume of both solute and solvent.

If this would matter on dilute solutions (~0.1 M), I cannot say, but nevertheless, the volume of solute in a solution is not that negligible.

Yes, basics in english.
Solution of course not solvent.
I work daily with analitical chemistry.

I added 4 gr of NaOH in 1 L of pure water

The proper way of making any mole SOLUTION is by adding for example those 4 grams of NaOH to graduated cylinder and filling it to calculated volume.
I'm making hundreds of such SOLUTIONS and didn't make any mistake for a year.

You can't include volume of NaOH because it changes.
It's like saying: adding 1L of water to 1L of ethanol gives you 2L SOLUTION

[Dacs]

Greetings!
That's correct fernandoz you used 0,1M (if you added 100% pure NaOH )
Since I'm a CHE student I think I'm closest to concentration matters.
Simple way to calculate concentration

c=n/V where n-number of mole, V-volume of solvent

n=m/M ,M for NaOH is 40g/mole

so for 0,1M=(4g/40g)/1L

including volume of NaOH is not permitted

I reached the desirable pH adding 20 ml of Soda Ash (.1 M Na2CO3)
If I use NaOH, then I'll be needing 30 mL .1 NaOH.

One thing you must check is purity of your soda ( NaOH is more basic than Na2CO3).
It's a common mistake to treat 70% NaOH as 100% pure one.

I think that you being a student just reinforces the fact that you need to review your basics.

Let me repeat, unless that I'm grossly wrong with what I know about concentration, molarity is defined as moles of solute / L of solution.

By solution, it means that you must account for the volume of both solute and solvent.

If this would matter on dilute solutions (~0.1 M), I cannot say, but nevertheless, the volume of solute in a solution is not that negligible.

Yes, basics in english.
Solution of course not solvent.
I work daily with analitical chemistry.

I added 4 gr of NaOH in 1 L of pure water

The proper way of making any mole SOLUTION is by adding for example those 4 grams of NaOH to graduated cylinder and filling it to calculated volume.
I'm making hundreds of such SOLUTIONS and didn't make any mistake for a year.

See item in bold

Do you honestly think that the water added that way will result exactly 1L?

Anyway, that's the proper way of preparing a solution, adding the solute first in a cylinder then adding enough water to make 1L of it. I think I've mentioned it in my 1st post here.

You can't include volume of NaOH because it changes.
It's like saying: adding 1L of water to 1L of ethanol gives you 2L SOLUTION

I certainly can't make any sense out of this statement.

All I'm trying to say is as per definition of molarity, you need to consider the volume of solution, not the solvent.

NaOH, even in solid form, occupies volume, you know

[Patrician]

Do you honestly think that the water added that way will result exactly 1L?

TO calculated volume
Let me see...I'm puting 4g of NaOH to graduated cylinder then filling it with water to 1L and hey the scale says 1000mL

You can't include volume of NaOH because it changes.
It's like saying: adding 1L of water to 1L of ethanol gives you 2L SOLUTION

you can't count the volume of NaOH being disolved because it changes.
Dissociaton occurs, crystal structure is destroyed.


Either we do not understand each other or you have to big ego to swallow "stupid student" reply.

[Dacs]

Do you honestly think that the water added that way will result exactly 1L?

TO calculated volume
Let me see...I'm puting 4g of NaOH to graduated cylinder then filling it with water to 1L and hey the scale says 1000mL

You can't include volume of NaOH because it changes.
It's like saying: adding 1L of water to 1L of ethanol gives you 2L SOLUTION

you can't count the volume of NaOH being disolved because it changes.

I did not say that you have to measure the volume of NaOH. All I'm saying is that you need to account for its volume. Adding exactly 1L of solvent will effectively lowers the concentration of the solution you're trying to prepare. That's what I've been saying all along.

And I think we agree on that, don't we?

Dissociaton occurs, crystal structure is destroyed.

Thank you for reminding me of the thermodynamics of mixing.

Either we do not understand each other or you have to big ego to swallow "stupid student" reply.

I think it's the former.

My point is, you made this post just a while back:

c=n/V where n-number of mole, V-volume of solvent

This is my contention, as someone who gives out replies to others, we have to make sure that we're giving out the right advices, otherwise it's just like blind leading the blind.

In your case, you have made the mistake of using the proper term (in which you acknowledged after). Consider this as lessons learned, try to proofread the reply before hitting the reply button.

[fernandoz]

Thanks everybody for your replies.

I have taken some notes, and they will help a lot.

Best regards,