# How to calculate the pH of solutions of amphoteric salts? [duplicate]

I've been trying to understand how to calculate the pH of solutions of amphoteric salts, but I just can't figure it out. So, I've seen that it is commonly explained like this (example for a solution of NaH2PO4):

The salt completely dissociates like so
$$\ce{NaH2PO4}$$-->$$\ce{Na+}$$ + $$\ce{H2PO4-}$$

Then the H2PO4- ion reacts with water to form HPO4 2- and H3O+. (reaction 1)
$$\ce{H2PO4-}$$ + $$\ce{H2O}$$ <=> $$\ce{HPO4 2-}$$ + $$\ce{H3O+}$$ _______ $$\ce{Ka2=6.2 x 10^{-8}}$$

And then it reacts with H3O+ to form H3PO4 and H2O. (reaction 2)
$$\ce{H2PO4-}$$ + $$\ce{H3O+}$$ <=> $$\ce{H3PO4}$$ + $$\ce{H2O}$$ ________ $$\ce{K=Ka1^{-1}, Ka1=7.52*10^{-3}}$$

The autoionization of water is negligible.

Now, I imagine this process like this:
The concentration of the $$\ce{H2PO4-}$$ ion in the first reaction is equal to the concentration of the salt in the solution because the salt dissociates completely. As it reacts with $$\ce{H2O}$$ to form $$\ce{HPO4 2-}$$ and $$\ce{H3O+}$$ (reaction 1), its concentration decreases for a certain value of x. The equilibrium concentration of $$\ce{H3O+}$$ and $$\ce{HPO4 2-}$$ in this reaction are the same and equal x. The equilibrium concentration of $$\ce{H2PO4-}$$ is equal to the concentration of the salt minus x.

$$\ce{[H2PO4-]1=c(salt)-x}$$
$$\ce{[H3O+]1=x}$$
$$\ce{[HPO4 2-]=x}$$

$$\ce{Ka2=\frac{[H3O+]1*[HPO4 2-] }{[H2PO4-]1}}$$

Then, what is left of the $$\ce{H2PO4-}$$, that is the equilibrium concentration of $$\ce{H2PO4-}$$ from the first reaction reacts with the equilibrium concentration of $$\ce{H3O+}$$ from the first reaction to form $$\ce{H3PO4}$$ and $$\ce{H2O}$$. The concentrations of $$\ce{H2PO4-}$$ and $$\ce{H3O+}$$ decrease for a certain value of y. So, in the equilibrium of this reaction, reaction 2, the equilibirum concentration of $$\ce{H3PO4}$$ will equal to y.

$$\ce{[H2PO4-]2=[H2PO4-]1-y=c(salt)-x-y}$$
$$\ce{[H3O+]2=[H3O+]1-y=x-y}$$
$$\ce{[H3PO4]=y}$$

$$\ce{Ka1^{-1}=\frac{[H3PO4] }{[H2PO4-]2*[H3O+]2}}$$

So the final concentration of $$\ce{H3O+}$$ is equal to the concentration of $$\ce{H3O+}$$ in the equilibrium of the second reaction.

$$\ce{[H3O+]final=[H3O+]2}$$

We can see that the final concentration of $$\ce{H3O+}$$ is equal to x-y. Because $$\ce{[HPO4 2-]=x}$$ and $$\ce{[H3PO4]=y}$$, we write:

$$\ce{[H3O+]final=[HPO4 2-]-[H3PO4]}$$

We can express $$\ce{[HPO4 2-]}$$ and $$\ce{[H3PO4]}$$ using the equilibrium constants, concentrations of $$\ce{H3O+}$$ and $$\ce{H2PO4-}$$.

$$\ce{[H3O+]final=\frac{Ka2*[H2PO4-]1 }{[H3O+]1}-\frac{[H2PO4-]2*[H3O+]final }{Ka1}}$$

If we assume that the initial concentration of $$\ce{H2PO4-}$$ is so large that the change of x and y are negligible, we can write that:

$$\ce{[H2PO4-]2=[H2PO4-]1=c(salt)}$$

If we assume that the concentration of $$\ce{H3O+}$$ is large enough so that the change of y is negligible, we can write:

$$\ce{[H3O+]final=[H3O+]1}$$

With those assumptions we can get this commonly used expression for explaining this problem

$$\ce{[H3O+]final=\frac{Ka2*c(salt)}{[H3O+]final}-\frac{c(salt)*[H3O+]final }{Ka1}}$$

and with a little bit of rearranging, we get this expression:

$$\ce{[H3O+]final=\sqrt{\frac{Ka2 * c(salt)}{1+\frac{c(salt)}{Ka1}}}}$$

What I don't understand is the fact that we were able to get the last equation because we assumed that the final concentration of $$\ce{H3O+}$$ ions is equal to their concentration in the equilibrium of the first reaction. But at the beginning, when we established the first equation for the final concentration of $$\ce{H3O+}$$, we assumed that the concentrations of $$\ce{H3O+}$$ ions in the equilibrium of the first and second(final) reaction aren't equal. If in the end we assumed they are equal, why can't we calculate the pH just using the first reaction? That way the $$\ce{H2PO4-}$$ wouldn't be treated as an amphoteric, which obviously isn't correct, so I'm wondering what is wrong in my approach?

• Jul 29 '20 at 0:35