Neutralisation Chemistry help required 1

[afran]

Some basic advice required for a non-chemical plant engineer.

Process Tank has 500 litre water to which is added 400 litre l5M NaOH followed by approx 600l flushing water. Tank is cooled prior to the addition of 12M HN03 acid and during process
What volume of HN03 would be required to neutralsise total volume to pH7.0?
What would be the molarity of the Sodium Nitrate produced in the final volume ?
Any calculation help most appreciated.

 

[25362]

To afran, although my chemistry is old history, I assume the following is about right.

[•] the neutralization would be exothermic, pH=7 is considered for a neutral solution at 25^oC;
[•] 1 mole of NaOH is stoichiometrically neutralized by 1 mole of HNO₃;
[•] the amount of moles of NaOH is 15 mol/L[×]400 L = 6000 moles;
[•] since the acid concentration is 12 mol/L the amount of acid solution needed would be: 6000 mol[÷]12 mol/L = 500 L.
[•] we have at the end 6000 mol of salt in about a 2000 L solution (assuming a density [≈] 1.0 kg/L)
[•] the final concentration would be 3M.

I hope I'm not mistaken.

 

[quark]

OK, two stupid questions from my side.

1. Should you not consider the 600L wash water?
2. Is the final volume not 1500 L (400+600+500) as per your calculation?
3.

 

[25362]

Original volume of water: 500 L
NaOH solution: 400 L
Flushing water; 600 L
Acid solution: 500 L

Total volume; 2000 L

The basic assumption being that the only "active ions" are those from the hydrolysis of NaOH and HNO3.

 

[afran]

Thanks 25362 for your explanation.That will allow me to do some further work on the dilution as am being asked some basic questions about limitation on the nitrate discharged to drains. The exothermic aspect is dealt with by having the process tank circulated through a Chiller plant to maintain vessel temperature during the operation.

 

[quark]

Yes, I got it. Just forgot that neutrilisation by dilution is a very poor technique. BTW, do you have any idea (empirical will do) about the molar concentration vis-a-vis pH for NaOH and HNO3?

 

[25362]

To quark,

Regretfully I don't have such tables. I assume a Google search may be of help .
However, I remember that an estimate of the pH at 25 Celsius of aqueous solutions of strong acids and strong bases (practically fully ionized) can be obtained as follows:

for the acid: pH = -log₁₀[acid]

example: [acid] = 0.2 M, pH = -log(0.2)= 0.7

for the base: pH = -log₁₀(1[×]10^-14/[base])

example: [base] = 0.2 M; pH = -log(10^-14[÷]0.2) = 13.3

 

[code1]

May I also add that it is important to see if the acid has 1, 2 or more hydrogen ions, e.g. HCL and H2SO4, etc.

For the above, a 0.2M solution of HCL and H2SO4 will give different pH's.

Please see also:

http://www.ausetute.com.au/phscale.html

 

[25362]

Thanks code1. the same adjustment should be made for bases. Let's repeat that the above applies for strong bases and acids which are considered fully hydrolysed.

For example 1 M of acetic acid at 25 Celsius is ionized only by about 0.42%.

Its K_a = 1.76 x 10^-5 = [H₃O⁺][Acetate^-][÷][acetic acid]

a good approximation would give [H₃O⁺]² = 1.76 x 10^-5 since the hydronium and the acetate anion are in equal proportions, and the acid, being weak and almost undissociated, is still at about 1 M.

[∴] [H₃O⁺]= 4.2 x 10^-3 M = 0.42% of the original concetration !

and the pH = -log (4.2 x 10^-3) = 2.38

When such an acid is more diluted the matter becomes more complicated because the acid becomes more ionized.

If its original concentration were 0.001 M,

then [H₃O⁺]=[Acetate]=1.24x10^-4 after solving a quadratic equation.

The fraction ionized is 12.4%, and the pH = 3.9.

Since the values of K_a and the initial concentrations may err by a few %, there is no need for using too many significant figures.

 

[code1]

Well, 25632, reading your post about weak acids (and weak bases)... just like school many years back.

Reading on the original post and early replies, I am just wondering, is there any need to consider Activity, instead of just concentrations, as the molarity is high.

 

[unclesyd]

What are you trying to get rid of the NaOH or the HNO3?

The dumping of the nitrate ion is considered as bad as dumping the acid in this part of the woods. If it's caustic the use of HCl is a better choice for neutralization as the end product is NaCl.

 

[25362]

To code1. It would be nice to know the individual ions' activity coefficients, but unfortunately these are not known. Thus, an additional error may be made in assuming concentrations as given.

The only way of doing things right would be by actually measuring the pH.

I just found a site that may be of interest:

http://www.nationmaster.com/encyclopedia/PH

to afran,

In addition to referring to the posting by unclesyd, can you kindly tell us whether an ulterior use of the nitrate solution as fertilizer or otherwise has been considered ?

 

[25362]

It is useful to remember that nitrate causes stress corrosion on mild steel.