These are the equilibria for protonated forms of PO43-.

H3PO4 H2PO4- + H+ HPO42- + H+ PO43- + H+

The pKas are, from left to right, 2.15, 7.2, and 12.32, according to one reference.

In 1 M KOH, the pH is considerably higher than 12.32, probably 14. Therefore, the phosphate will be mostly in the PO43- form, with a small amount (~2%) protonated into the HPO42-. Since this consumes protons, you might expect the pH to go up slightly instead of down. However, you observed the pH go down slightly, so I suggest that the decrease in pH was due to the dilution of KOH caused by increasing he volume of the solution. The volume increases because of the solid Na3PO4 takes up space once it is dissolved. What was the volume of the solution after the Na3PO4 was added? What happens to the pH if you increase the volume of the solution the same amount by adding water?

We must also consider the suggestion of Raja Summe Ullah . I saw a suggestion that NaOH is actually a stronger base than KOH. If so, I don't think that is the solution.

Are you sure it was trisodium phosphate, not monobasic sodium phosphate or dibasic sodium phosphate?

Please mention all the conditions of your addition. How much pH was decreased? Did you observe some reaction or heat changes during it?

Adam B Shapiro Yes, its Na3PO4.

Raja Summe Ullah I just mixed them as mentined at room temperature, no obvious heat change was found. And the pH decreased about 0.15.

I think the slight pH decrease can be attributed to the fact that Na3PO4 is a weaker base than KOH, so that adding it as a solid to the KOH solution was essentially the same as diluting the KOH a little.

Is this not what you would expect. 50ml of 1M KOH is equivalent to 2.8 g of KOH while 2g of phosphate with an annhydrous MWt of 163 in molar terms is about 1/3 the amount of KOH so even given that phosphate is a weaker acid it should have some effect over and above the dilution factor

Adam B Shapiro In my view, the hydrolysis of PO43-, PO43- + H2O → HPO42- + OH - , leads to the formation of OH-, so the pH shold be constant or increase slightly.

Paul Rutland In my view, the hydrolysis of PO43-, PO43- + H2O → HPO42- + OH - , leads to the formation of OH-, so the pH shold be constant or increase slightly.

The pKa of HPO42- is 12.4. At the pH of 1 M KOH, probably about 14, there will be very little HPO42-. The equilibrium will favor PO43-.

I agree with Pr. Adam B Shapiro.

Thanks a million,

Adam B Shapiro Thank you for your helpful answer. I am still confused. The existence of HPO42- indicates the hydrolysis of PO43- and the formation of OH-, thus higher OH- concentration and higher pH?

I also agree with Pr. Adam B Shapiro. One more possibility could be the replacement reaction between Na and K which may result in the formation of NaOH, even in small quantities. Which is slight weaker than KOH.

These are the equilibria for protonated forms of PO43-.

H3PO4 H2PO4- + H+ HPO42- + H+ PO43- + H+

The pKas are, from left to right, 2.15, 7.2, and 12.32, according to one reference.

In 1 M KOH, the pH is considerably higher than 12.32, probably 14. Therefore, the phosphate will be mostly in the PO43- form, with a small amount (~2%) protonated into the HPO42-. Since this consumes protons, you might expect the pH to go up slightly instead of down. However, you observed the pH go down slightly, so I suggest that the decrease in pH was due to the dilution of KOH caused by increasing he volume of the solution. The volume increases because of the solid Na3PO4 takes up space once it is dissolved. What was the volume of the solution after the Na3PO4 was added? What happens to the pH if you increase the volume of the solution the same amount by adding water?

We must also consider the suggestion of Raja Summe Ullah . I saw a suggestion that NaOH is actually a stronger base than KOH. If so, I don't think that is the solution.

Adam B Shapiro Thanks a lot for your kind help!

Thanks a million Dear Adam B Shapiro for these explanations.

Adam B Shapiro Thank you so much Prof for this correction and for detailed answer.

Dear Feng Zeng, your idea that pH should increase is right. When trisodium phosphate dissolves, all phosphate species and even phosphoric acid appear in solution. This occurs through hydrolysis equilibria that generate OH- ions. Of course, the amount produced by these equilibria will be limited by the effect of the common ion due to the KOH initially present, but necessarily, it should increase the pH even slightly. The problem is not the chemistry involved but the limitations of the instrumental probe used to measure pH.

The decrease in pH you have observed is probably due to a problem already known about the alkaline error experienced by the combined glass electrodes to measure very high pH. A detailed explanation can be found in classic "Instrumental Analysis" texts. Anyway, I'll try to summarize here an answer.

The pH electrodes have a porous glass membrane especially sensitive to H+ ions. In strongly alkaline solutions, the few H+ ions have to compete with other cations that are in abundance, as is the case you raise: K+ and Na+. The Na+ is relatively small and at a relatively high concentration (in your case 0.73 M approximately) it can permeate the membrane to a certain degree by stimulating a response in the electrode that is computed as a greater presence of H+ ions, as a result, a slight decrease in pH occurs.

i agree with prof Adam

It is evident that the pH measurements are correct if taken with a previously calibrated instrument. The expected result of the calibration is a straight line that relates the instrumental (potential) response to the pH. The slope of this line is constant in a wide pH range but it shows deviations towards the extreme pH, such deviations are referred to as the acid error and the alkaline error.

The alkaline error, also known as sodium error, causes a decrease in the actual readings, which possibly happened to Feng Zeng. This error can be of greater or lesser magnitude depending on the type of sensitive glass used in the pH specific electrode. Below is an interesting article that explains this, review the text on page 211 and figure 1 (on the same page) that clearly illustrates the alkaline error (and also the acid error) of different brands of glass electrodes for pH .

I am going to share this question with one of the authors of that article, which has an account in ResearchGate, maybe he wants to join us in this discussion.

Another aspect I wanted to comment on is the acid-base equilibria that are really established in the solution prepared by Feng Zeng. Upon dissolution of Na3PO4, all possible phosphate species (HPO42-, H2PO4- and H3PO4) were generated in the solution by equilibrium reactions known as hydrolysis equilibria. These are the following:

PO43- + H2O HPO42- + OH-

HPO42- + H2O H2PO4- + OH-

H2PO4- + H2O H3PO4 + OH-

At the beginning before these reactions occurred, there was already a large initial amount of OH- ions coming from the KOH that would affect the position of these equilibria (effect of the common ion). Then these reactions moved from left to right until equilibrium was reached, this would necessarily imply that more OH- ions would be formed than were originally available. Now, will this mean that the concentration of OH- would increase? In this discussion it has been suggested that the dissolution of Na3PO4 would have decreased the concentration of OH- by a dilution effect and consequently the pH would have dropped. This hypothesis could be revised by making equilibrium calculations. In the following comment I will present my estimates in this regard.

Postscript: the following equilibria did not operate in the solution:

H3PO4 + H2O H2PO4- + H3O+

H2PO4- + H2O HPO42- + H3O+

HPO42- + H2O PO43- + H3O+

Why not? The reactions would have to move from right to left and a supply of H3O + would be needed for part of the water that is not thermodynamically possible, without breaking the water's own equilibrium constant, Kw.

Dear colleagues, after making the calculations, I do not consider probable the hypothesis that the decrease of the pH of 0.15 units observed by Feng Zeng has been by dilution. I enclose the Excel document in which I carry out all the necessary calculations. These calculations show a decrease of only 0.008 pH units (from 14 to 13,992) considering some extreme assumptions for simplicity. More realistic considerations would make the aforementioned hypothesis even less likely.

Please note that:

I) For PO43-, anion hydrolysis writes: PO43- + H2O ⇌ HPO42- + OH- (i); with KhPO43- = [HPO42-]·[OH-]/[PO43-] = Kw/Ka3H3PO4 = 2.4·10-2, where Ka3H3PO4 stands for the 3rd dissociation constant of H3PO4 and Kw = 1.00 x 10-14; units were omitted. Species H2PO4- and H3PO4 can be safely neglected for the mentioned alkaline solution.

II) At the relatively concentrate solution of the strong base KOH; equilibrium (i) becomes largely displaced to the left, hence consuming OH- and HPO42- from the solution.

III) Although the Na3PO4 salt can be predicted to possess weak basic character at its own isolated aq. sol., it behaves as acidic at the fairly concentrate solution of the much stronger base KOH.