Dihydrogen Phosphate has a Ka of 6.2x10^-8, yet it seems to form solutions with an acidic pH when it is in water. Why is that? Shouldn't it dissociate back to phosphoric acid and form a basic solution, since its Ka is less than 1.0x10^-7?

It seems as though Ka does play a role in determining whether an amphoretic substance will act like a base or acid, since 0.10 M Hydrogen Phosphate has a Ka of 4.8x10^-13 and creates a basic solution. However, I feel like I am missing another mechanism that determines the behavior of an amphoretic substance in a neutral solution.


2 Answers 2


The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species $\ce{H2PO4-, HPO4^{-2} , PO4^{-3} ,H3PO4}$ in solution, that ($\ce{H2PO4-}$) is amphoteric that behaves both:

as an acidreleasing $\ce{H+}$ in aqueous solution and( forming $\ce{HPO4^{-2}}$ ions which can produce very few of $\ce{PO4^{-3}}$ ions)

$$\text{ Ionization:}\ce{H2PO4- <=> HPO4^{-2} +\color{red}{H^+} \quad\left(K_\mathrm{a}=\color{red}{6.2\times10^{-8}}\right)}$$ OR

as a base(can undergo hydrolysis forming ($\ce{H3PO4}$) acid $$\text{Hydrolysis:}\ce{H2PO4- + H2O <=> H3PO4 + \color{blue}{OH-} \quad\left(K_\mathrm{b}=\color{blue}{1.3\times10^{-12}}\right)}$$

So ,$\color{red}{K\mathrm{a}}>\color{blue}{K\mathrm{b}}$ ,thus the first reaction is the dominant,so ,$\pu{pH}<7 $ and the solution is$~\color{red}{\text{acidic}}$.


You don't really compare the $K_a$ value with the extent of autodissociation in pure solvent. Rather, for an sacid compare the generation of solvated hydrogen ions by the solute dissociation versus solvent (water) dissociation.

Here, solvent dissociation would give up to $10^{-7}\text{ M}$ solvated hydrogen ion.

Compare this with the solvated hydrogen ion coming from the dihydrogen phosphate ion. Letting $x$ be the amount of solvated hydrogen ion that would be formed this way you have:


where you use the stoichiometry based on the reaction $\ce{H2PO4^-<=>H^+(solv) + HPO4^{2-}}$. When you solve the quadratic equation for $x$ you will get $7.9×10^{-5}\text{ M}$, which overwhelms the maximum autodissociation by a factor of almost eight hundred. In fact all this excess solvated hydrogen ions make a negative feedback, cutting down the autodissociation more via Le Chatelier's Principle. Your solvated hydrogen ion concentration is dominated by the solute, so the solute registers as an acid.

  • $\begingroup$ Solving the same equation for hydrogen phosphate also gives you a value for x that is over 1.0x10^-7, which is 2.12x10^-7, yet it tends to hydrolyze backwards. $\endgroup$
    – chemN00b
    Jan 8, 2019 at 0:31
  • $\begingroup$ Be sure you have the amount of monohydrifen phosphate ion actually formed in the first dissociation, use that for the second dissociation equilibrium. $\endgroup$ Jan 8, 2019 at 0:47
  • $\begingroup$ Sorry I wasn't clear enough. I meant that if you put in 0.1M sodium hydrogen phosphate in neutral water, it would still produce a greater H+ ion concentration than water would. $\endgroup$
    – chemN00b
    Jan 8, 2019 at 0:55
  • $\begingroup$ Next comparison -- use the Kb value to see how much hydroxide comes from the monohydrogen phosphate ion. Which dominates now? $\endgroup$ Jan 8, 2019 at 1:08
  • $\begingroup$ Ah, I see. However, I'm still confused. Plugging in the Kb value for dihydrogen phosphate gives me a greater concentration of OH- than H+, with [OH-] = 1.2x10^-4. I used the equation X^2/(0.1-x) = 6.2x10^-8/10^-14 $\endgroup$
    – chemN00b
    Jan 8, 2019 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.