Hi,
a friend of mine found the following question in preparation tests for medicine:
How much water do you have to add to 1 ml of solution with pH 8 to reach pH 5?
I assume they want simply calculation pH = -log [H+]
But in reality, what would be the pH of for example NaOH with concentration e.g. 10^-9 M? In such case the concentration of OH- originating from the hydroxide would be 10^-9 M, so at first sight, one could say the pH would be pH = 14 - pOH = 14 - (-log(10^-9)) = 14 - 9 = 5.
But of course the hydroxide will not have acidic solution, because the water dissociates. Is there a (simple) way to calculate the pH?
Just to make sure - that question is nonsense, right?
Thank you
https://socratic.org/questions/how-do-you-calculate-the-ph-of-a-solution-when-given-the-oh-concentration
https://sensorex.com/ph-calculator/
The pH of 0.000000000001 M sodium hydroxide is 7.
Yes, I know that hydroxide is base and hence the pH shall be above 7. That's why I'm asking.
But actually the video you linked is providing the same answer that the authors of my question probably wanted to hear:
https://youtu.be/gn1CgBzShps?t=541
So there the Mg(OH)2 solution is acidic as well.
Note that close to neutral pH (7) we can not neglect [H3O+] compared to [OH-]. Moreover, the contribution of CNaOH = 10-9 M for the pOH is two orders of magnitude smaller than that from water when pH ≈ 7: [OH-] = 10-7 M.
For CNaOH = 10-9 M, it is quite reasonable to simply neglect the contribution of NaOH for the pH, because it is utterly small. The pH is essentially contributed just by water.
Nevertheless, the pH can still be calculated accounting for both the water and NaOH contribution; as follows.
Sodium molar balance writes: CNaOH = [Na+] + [NaOH], where CNaOH is the nominal molar concentration of NaOH in aq. sol. (formality, to be precise). Because NaOH is a strong base, it should be highly dissociated in aq. sol., so that [NaOH]
Thank you Carlos Araújo Queiroz! I knew it has to be something like this, but didn't know how to calculate it.
Thank you once again!
I'm just looking now at the calculation. What is the M at the end of the calculations? Is that units? Like in pH = - log10{[H3O+]/M} ?
About predicting the pH of Ca(OH)2 aq. solutions (saturated or not) ― cf. my posts (§ III., IV, V.) at:
https://www.researchgate.net/post/How-to-increase-pH-for-a-solution-without-making-a-chemical-reaction-with-CaCl2
On pH prediction for Cu(OH)2 aq. sol.:
https://www.researchgate.net/post/Is_it_possible_to_ppt_out_CuOH2_at_acidic_pH_4-6
On pH prediction for KOH aq. sol. ― cf. my posts (§ I, II.) at: https://www.researchgate.net/post/How_to_Prepare_1_M_KOH_solution_with_85_KOH_pellets
On pH prediction for LiOH aq. sol.:
https://www.researchgate.net/post/What-is-the-pH-of-lithium-hydroxide-in-water
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