I recently heard someone claim that adding table salt to vinegar caused HCl to form which helped them clean coins - clearly not the case - however I thought I'd disprove it by doing an experiment.

So I added NaCl to ethanoic acid and measured a 0.5 reduction (3 to 2.5) in pH using a pH probe. This seemed a bit weird so I repeated using tap water and the same happened (7 to 6.5).

The pH of water when NaCl is added should not change from 7!

The probe measures by detecting a change in voltage through the solution so it could be that, but you'd think that salts shouldn't have such an effect or be adjusted for. Can anyone repeat please?

What's going on?

  • $\begingroup$ If you add NaCl to any neutral or acidic solution, the pH will decrease by a factor of at least 0.5. This is due to the fact that the pH is not the log of the concentration calculated with respect to the total volume, but the concentration should be calculated with respect to the volume of free water. In concentrated NaCl solutions, nearly all water molecules are adsorbed around Na+ and Cl- ions. The apparent concentration of H3O+ increases a lot for lack of water. $\endgroup$
    – Maurice
    Commented Nov 10, 2019 at 14:54
  • $\begingroup$ I know salt lowers the measured pH because when I sprinkle a tiny bit of salt into distilled water I can watch my pH meter drop to a reading of 6.4 so it does affect the pH meter reading whether or not it actually changes the true pH. $\endgroup$
    – George
    Commented Apr 29, 2023 at 0:11
  • $\begingroup$ First: the experiment is lacking in details, Tapwater exactly 7!! The data presented lacks precision. If both readings changed 1/2 a unit that is 30mv, That is a large difference in a[H+]in the two solutions. That makes the reference electrode junction potential suspect. Another possibility is the actual change in Kw and Keq of acetic acid. First ensure the equipment is measuring appropriately. $\endgroup$
    – jimchmst
    Commented Apr 29, 2023 at 2:08
  • $\begingroup$ I’m voting to close this question because it lacks information about the type of electrode, the calibration procedure and the definition of pH at high ionic strength, see e.g. core.ac.uk/download/pdf/38093394.pdf $\endgroup$
    – Karsten
    Commented Apr 29, 2023 at 3:16

2 Answers 2


This is probably an effect in which sodium ions from the salt are being weakly detected by the glass electrode, giving the appearance of a lower pH. Try add the same concentration of salt to samples of your reference buffers and recalibrate your pH meter to those before measuring the pH of the salt water. I would expect you would see pH 7. Also, keep in mind that the pH of unbuffered water is very sensitive to low concentrations of certain impurities -- from the air or from the added salt. For example, carbon dioxide in the air will lower the pH over time.

  • $\begingroup$ If this were true, a basic solution would also shift towards lower pH after adding salt. Arguments involving ionic strength would predict the opposite. These predictions are easy to test experimentally - add some sodium chloride to a solution of a weak base and compare to adding water, making sure the final volume is the same. Then, measure pH values and compare. $\endgroup$
    – Karsten
    Commented Nov 11, 2019 at 3:16
  • $\begingroup$ I think that the response of the glass electrode to sodium ions is more important than the activity coefficient changes. $\endgroup$ Commented Jan 6, 2020 at 9:24
  • $\begingroup$ Modern electrodes have a minimal Na+ or other monovalent ion sensitivity. $\endgroup$
    – jimchmst
    Commented Apr 29, 2023 at 1:42

Please be aware that adding salt changes the ionic strength of your solution. Therefore, the activity $a_\mathrm H$ of $\ce{H+}$ is changed. This changes $\mathrm{pH}=-\log_{10}(a_\mathrm H)$. In a 1 molar $\ce{NaCl}$ solution ($I\approx 1\ \mathrm{mol/l}$) the Debye-Hueckel (DH) approximation gives a shift of the activity by around 0.5 pH units (compared to salt free water):

The activity coefficient $f$ is given by DH

$\log_{10}(f_i) \approx - A \sqrt{I/(1\ \mathrm{mol/l})}$, with ionic strength $I$ and $A=0.509$

This gives $\mathrm{pH}=-\log_{10}(c_\mathrm H/(1\ \mathrm{mol/l})) -\log_{10}(f_\mathrm H)=-\log_{10}(c_\mathrm H/(1\ \mathrm{mol/l}))+A \sqrt{I/(1\ \mathrm{mol/l})} $.

This equation would predict an increase of pH by 0.509 pH units for a 1 molar salt solution.

You might want to use a better theory than Debye-Hueckel. In the above form it is only applicable for ionic strengths below $\sim 0.001\ \mathrm{mol/l}$, so the calculation above is very crude and might explain the failure.

Alternative explainations are given by https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3638298/

  • 1
    $\begingroup$ +1 for mentioning activity coefficients and ionic strength. However, if the activity coefficients of hydrogen ions and acetate ions are different from one, the position of equilibrium has to be recalculated (more acetic acid dissociates). This reference gives such a calculation for a different weak acid/base system at ionic strength 0.1 M. In that example, the pH of the solution changes marginally towards lower pH. $\endgroup$
    – Karsten
    Commented Nov 11, 2019 at 3:08

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