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I am an engineer in water treatment and have a chemistry question. In water treatment we dose $32\%$ sodium hydroxide solution for $\mathrm{pH}$ adjustment and there is a concern that our dosin lines must be $100\%$ dry before introducing the $\ce{NaOH}$ solution. This is the problem as they have been pressure tested with water and there is water residue in them.

Will $32\%$ sodium hydroxide solution react violently with cold water? It is my understanding that the reaction only occurs when solid sodium hydroxide (lye) is dissolved in water and since my $\ce{NaOH}$ is already in solution then this reaction will not take place. Is this correct?

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The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance.

What can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite diluted, I expect temperature change rather few than many degrees.

Another concern is initial dosage would be affected by water, decreasing concentration of the applied volume.

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  • $\begingroup$ Thanks for your help, I am not a chemist so I am unable to work out the exact release of heat in the reaction and the amount of residual water present would be impossible to quantify. i can say that the longest dosing line has a volume of 14 litres when full but of course this will have been emptied prior to introduction of NaOH. The worst case possible is that 14 litres of water mixes with 7000 litres of NaOH. This will never happen but at least I would know that my scenario will generate far less heat. Is this something you would be able to quantify at all? $\endgroup$ Jun 18 '20 at 7:56
  • $\begingroup$ I mean taking a small volume of solution , diluting with water and measure temperature raise. It can be done without lab equipment. But unless it is PVC tubing with low melting point, I would not be concerned. $\endgroup$
    – Poutnik
    Jun 18 '20 at 7:58
  • $\begingroup$ With such volume ratio, I would be of no concern. $\endgroup$
    – Poutnik
    Jun 18 '20 at 8:04
  • $\begingroup$ It is PVC pipework $\endgroup$ Jun 18 '20 at 8:04
  • $\begingroup$ I didn't think so, thanks very much for your help $\endgroup$ Jun 18 '20 at 8:05
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I'd not concern about the heat increased by the dilution of $32\%$ $\ce{NaOH}$ with negligible amount of water. I'm telling this by my personal experience. In our organic lab program, our student workers have been diluting $50\%$ $\ce{NaOH}$ solution ($\pu{12.5 M}$) with water at room temperature to make $\pu{3 M}$ ($\ce{NaOH:H2O}$ is in $1:3.2$ ratio) and $\pu{6 M}$ ($\ce{NaOH:H2O}$ is in ~$1:1.1$ ratio) solutions in $\pu{20 L}$ capacity. These ratios are worse than the ratio in your worst case scenario ($\ce{NaOH:H2O}$ is in $1:0.002$ ratio), assuming $\ce{NaOH}$ percentage is in $w/v$. None of dilution has reached untouchable (to my skin! :-)) temperatures. This means, the temperature increment would be far below the melting point of PVC tubing in your worst case scenario. Yet, since this is just based on personal experience, I'd like to give you some scientific evidence.

In Caustic Soda Handbook of OxyChem, there is an example saying "the approximate resulting temperature of final solution after diluting $50\%$ $\ce{NaOH}$ at $\pu{120 ^\circ F}$ ($\pu{48.9 ^\circ C}$) to $20\%$ using water at $\pu{80 ^\circ F}$ ($\pu{26.7 ^\circ C}$) is $\pu{143 ^\circ F}$ ($\pu{61.7 ^\circ C}$)." This would result $\pu{63 ^\circ F}$ ($\pu{35 ^\circ C}$) raise with respect to water temperature. The Graph 6: Relative enthalpy of aqueous caustic soda solutions (PDF) in page 36 would provide you with some important insights. The similar scenario you are facing is described here as well.

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