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NaOH dilution/pH help

NaOH dilution/pH help

NaOH dilution/pH help

(OP)
I am trying to relive my first year chemistry at work today and it's not treating me very well.  

I have a situation where I'm using a 20% wt solution of NaOH.  I need to provide 325 mol (13 kg) of NaOH (so 65kg of solution).  The density I found for NaOH at 20% is ~1.2 kg/L leaving me with ~55L.  I come up with a pH of 14.7 for this solution.  If I remember correctly a pH of 14 is possible.

I then need to find how much water I need to dilute the solution to a pH closer to 7, so lets say 7.  The figure I am getting is very very large and seems unrealistic to me.  Can someone please help me out.  

I have checked eng-tips and google and this question seems straighforward but I cannot seem to make it work.

Cheers.

RE: NaOH dilution/pH help

13 kg of NaOH would be 0.325 mol but above you say you need 325 mol which would mean you would need 13,000 kg. You may want to check your numbers.

RE: NaOH dilution/pH help

Pure water has a pH of 7 because it has equal number of H+ and OH- (if I remember correctly it is 10-7 mole/lit)  If you had NaOH, there will theoretically always be more OH- ions than H+ ions so you will never reach pH of 7.

It has been a while since chemistry, but I think this is how you do it.

I think you need to calculate the moles added of NaOH based on the molecular weight and calculate the liters of water.  Since NaOH is has only a single OH-, 1 mole of NaOH will equal 1 mole of OH- ions.  That would give you the m/l of OH- in solution.

pH = 14-log(base 10)of moles per liter of OH-

That should give you your pH (I think).

RE: NaOH dilution/pH help

Macmet,

pH of ANY solution cannot be higher than 14. For high concentrated solutions, the calculation of ionic strength has to be employed. It means that not all of the ions present in the solution will exist as separate species; when electrolyte concentration becomes too high (above certain limit), these ions will recombine again and give electroneutral specie (initial compound).

About dilution water quantity: pH is a logarythmic value. Just as an illustration - for decreasing pH from 14 to 7 by using pure neutral H2O, you would need approx. 10,000,000 m3 of water for one 1m3 of alkaline solution. Is the picture more clear now?

Regards,

http://antwrp.gsfc.nasa.gov/apod/astropix.html

RE: NaOH dilution/pH help

(OP)
Ok,

I need 325 mol of NaOH.  Molar mass of NaOH = 40g/mol.

325 mol * 40 g/mol = 13,000 g = 13kg.... unless I'm way off.

Pedarrin, the method you outlined is what I attempted.  I'm looking at my figures and I may have a negative sign wrong so I will go through them again, but I have to get to a meeting.

RE: NaOH dilution/pH help


pH values above 14 and below zero are possible.

RE: NaOH dilution/pH help

" I come up with a pH of 14.7 for this solution.  If I remember correctly a pH of 14 is possible."


Where do you come up with a pH of 14.7? That is impossible since the pH scale ends at 14.

 I come up with a pH of 14.7 for this solution.  If I remember correctly a pH of 14 is possible.

http://www.dow.com/PublishedLiterature/dh_0048/0901b803800483fc.pdf?filepath=causticsoda/pdfs/noreg/102-00484.pdf

"I then need to find how much water I need to dilute the solution to a pH closer to 7, so lets say 7.  The figure I am getting is very very large and seems unrealistic to me.  Can someone please help me out."

This is also impossible. Pure water will not neutralize caustic. A pH of 7 indicates that the solution is neutral. You would have to add and acid to lower the pH of caustic to 7. See the dow file above for an estimate of the diluted pH.

  
Your numbers look correct as to how much caustic is required, but your concept of the pH scale is off target.



RE: NaOH dilution/pH help


Pure and neutral water at 60oC has pH=6.5. At this temperature a pH=7 would indicate that water is alkaline.

[H3O+][OH-] = Kw thus [H3O+] = Kw1/2

Kw at 25oC = 1.01 × 10-14→pH = 7
Kw at 60oC = 9.61× 10-14→ pH = 6.5

RE: NaOH dilution/pH help


From the internet

Quote:


Autodissociation of water can be a controlling factor in certain regimes...but definitely not in the strong acid / strong base regime.
As it turns out, commercial concentrated HCl (37% by weight)
has a pH of approximately -1.1....and saturated NaOH
solution has a pH of about 15.0 [source: Chemical Principles, 4th edition, by Dickerson/Gray/Darensbourg/Darensbourg].

RE: NaOH dilution/pH help

Thanks to 25362 for a more precise definition of pH. Let me expand the concepts further.

pH is a measuurement of H+ ion concentration.

In aqueous solutions, pure water has a neutral pH of 7.0. That is equivalent to having a presence of 10exp(14-7.0) H+ ions and 10exp(14-7.0) OH- ions. Since H+ concentration equals OH- concentration, the solution is neutral.

When some acid is added to pure water, the measured pH decreases below 7.0. The value of pH is a function of concentration, ability to dissociate (strong to weak), and temperature. As H+ concentration increases, pH value decreases 10exp(14-6.0) equals 10exp8.

When some base is added to pure water, the measured pH increases above 7.0. The value of pH is a function of concentration, ability to dissociate (strong to weak), and temperature. As OH- concentration increases, pH value increases 10exp(14-8.0) equals 10exp6. There is an of OH- ions.

Dilution of an acidic or basic solution with pure water decreases the concentration of the solution. The apparent pH measurement will change slowly towards an asymptotic 7.0. Reaching 7.0 by dilution requires an infinite amount of pure water. It is not practical.

Neutralization of an acid with a base or a base with an acid will yield pH of 7.0. It is a more common approach. As a simplified concept, a mole of acid to neutralize a mole of base.

In industry, common practice is to have a target pH range that needs to be achieved. As an example pH of 7.2 to 7.8.











RE: NaOH dilution/pH help


The pH is defined as the log10 of the inverse of the activity of the hydronium ion. The activity itself is the product of the ion's concentration multiplied by a correcting factor named coefficient of activity.

The CRC Handbook (77th Ed.) gives for NaOH the following values for the coefficient of activity:
 
molality (mol/kg water)     coefficient of activity

           0.001                           0.965
           0.100                           0.775
           1.000                           0.674
           5.000                           1.076
           10.00                           3.258
           20.00                           19.41





RE: NaOH dilution/pH help

Ok - you are talking g mol. That is something you hadn't defined. It could be a lb mol, kg mol or as now you have said g mol.

RE: NaOH dilution/pH help

pH cannot be higher than 14.
Come on friends, chemical engineering literature philosophy is a way without end. Here, we are talking about practical issues and concerns.

25362, do you believe that pH can be higher than 14?

http://antwrp.gsfc.nasa.gov/apod/astropix.html

RE: NaOH dilution/pH help

I apparently have missed something. If MacMet's original concern is to form a water solution of pH 7, why is he concerned with the dilution of a NaOH solution. Why not simply use water unadulterated except for some minor alkaline or acid addition to achieve pH 7?

Orenda

RE: NaOH dilution/pH help

(OP)
Orenda,

I just randomly picked a pH of 7 hoping someone would help me figure out how much water I would need to dilute the solution to that point.  It turns out I should have picked a value like 10 or 11.  If someone still wants to show me how to calculate how much water I'd need I'd appreciate it.  I think I know what to do now but I haven't had the chance to try yet.  

In regards to my original 14.7, I checked the numbers and I had a negative mixed up in there and I now have ~13.3.  I'm comfortable with that # b/c it is similar to what DOW shows in the link provided by BIMR.

Thanks for everyone's help, I will keep monitoring any posts.

RE: NaOH dilution/pH help


I'd be cautious about extrapolation from 3% mass (0.77 M), the last point in Dow's graph, to 20% mass (~6.1 N), macmet's case, due to the appreciable increase of the activity coefficient.

Most textbooks (probably wrongly) ascribe a pH≈14.0 to a 1N NaOH (about 3.9% mass) solution.

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