Dear colleagues, what is the pH of lithium hydroxide in water? And can it be used for geopolymerisation process??
LiOh is not a weak base, meaning it will dissolve completely in water. To calculate the pH of a solution of LiOH you only need to know the concentration as Myron points out.
In addition to suggestions from Myron and Sigurður, you can use the following equations to estimate the pH:
pOH = - log [OH-]
pH + pOH = 14
For geopolymerisation process you have to specify what exactly you want to do. There are examples here:
Article The geopolymerisation of alumino-silicate minerals
Thank you dear researchers,
I wanted to know whether lithium hydroxide can be used in place of sodium hydroxide in polymerization of concrete???
It will be very helpful to know your suggestions
Regards
Abhijeet Baikerikar
Dear Abhijeet Baikerikar ,
I am not familiar to your work, but talking about the difference between the alkalinity difference between LiOh and NsOH only. Although there is no strong agreement about the real difference. However, some have concluded that LiOH is slightly less IONIC tnan NaOH. Therfore NoOh gives slightly higher pH than LiOH. If you calculate pH of 0.1M each of LiOH and NaOH, giving that pKb of LiOH=-0.36 and pKb for NaOH=0.2, although these values are not completely accurate, but anyway, accodingly the NaOH pH is about 0.1 pH unit higher (means more alkaline) than LiOH at o.1M concentation. You may search more in the literature if you need higher certinity. Thank You and GOOD LUCK .
Dear Abhijeet Baikerikar ,
I am writing to you again just to apologize for the mistakes I dis about the correct structure of both LiOH and NaOH in my answer to your question yesterday. Because I did it quite rapid without its revision, I just saw the mistakes just now. Sorry once again, and wish you a GOOD LUCK
The pH of LiOH aq. sol. can be predicted from the hydrolysis constant (Kh = 2.5·10-14; units omitted) (*) of Li+ for the equilibrium: Li+ + 2H2O ⇌ Li(OH) + H3O+; Kh = [Li(OH)]·[H3O+]/[Li+]. We can substitute [Li(OH)] at previous equation based on charge balance and molar balance for lithium, respectively: [Li+] = [OH-] - [H3O+] = Kw/[H3O+] - [H3O+], [LiOH] = CLiOH - [Li+] = CLiOH - [OH-] + [H3O+] = CLiOH - Kw/[H3O+] + [H3O+], where Kw is autodissociation constant of water (Kw = 1.0·10-14; units omitted) and CLiOH is the nominal molar concentration of dissolved LiOH (formality, to be precise). Hence we can write: Kh = {CLiOH - Kw/[H3O+] + [H3O+]}·[H3O+]/{Kw/[H3O+] - [H3O+]} = {{[H3O+] + CLiOH}·[H3O+] - Kw}·[H3O+] /{Kw - [H3O+]2}. As long as CLiOH > 100·[H3O+] and CLiOH > 100·Kh, this simplifies to: [H3O+]2 - (Kw /CLiOH)·[H3O+] ≈ Kh ·Kw/CLiOH. This yields:
[H3O+] ≈ (½)·(Kw/CLiOH)·{1 + √((4·CLiOH/Kw) + 1/Kh)}·√Kh.
For instance; for 0.1 M and 1 M aq. sol., we predict pH =13.15 and 13.78, respectively.
(*) pKa = - log10(Kh) = 13.6; cf.: https://www.wou.edu/las/physci/ch412/Table2.htm
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