Find the pH of an amphiprotic salt dissolved in water at room temperature

Let's say I have a polyprotic acid $\ce{H2A}$ with the following properties:


$$
\begin{align}
\ce{H2A &<=> HA- + H+} & K\_1 &= 1\cdot 10^{-1}\\
\ce{HA- &<=> A^2-+ H+} & K\_2 &= 1\cdot 10^{-4} 
\end{align}
$$


Now a 0.5 M solution of $\ce{HA-}$ is prepared; what will be the $\mathrm{pH}$ of the solution?


My textbook (and many other sources) says it should be


$$\frac{\mathrm{p}K\_1 + \mathrm{p}K\_2}{2}$$


which sounds a little counterintuitive to me.


First of all, it implies that the $\mathrm{pH}$ is independent of the concentration of $\ce{HA-}$ present.


Also, if $1/\mathrm{p}K\_1 > \mathrm{p}K\_2$, then $\ce{HA-}$ should have the tendency to accept a $\ce{H+}$ more than it has of losing one, then why is the $\mathrm{pH}$ exactly partway between $\mathrm{p}K\_1$ and $\mathrm{p}K\_2$?




$$\begin{align}
\mathrm pH &=\mathrm
pK\_\mathrm{a1}+\log \frac {[\ce{HA-}]} {[\ce{H2A}]} \\
\mathrm pH &=\mathrm pK\_\mathrm{a2}+\log \frac {[\ce{A^2-}]} {[\ce{HA-}]} \\
\mathrm pH &=\frac{\mathrm pK\_\mathrm{a1} + pK\_\mathrm{a2} +\log \frac {[\ce{A^2-}]} {[\ce{H2A}]}} {2}\\
\end{align}$$


The last equation is the sum of the equation sides, divided by 2. 


We can consider for expected acidic $\mathrm pH$, that we can neglect water dissociation and suppose $\ce{H+}$ is produced and consumed by these 2 reactions:
$$\begin{align}
\ce{HA- + H+ &<=> H2A} \\
\ce{HA- &<=> H+ + A^2-} \\
\end{align}$$


Therefore:


$$\begin{align}
[\ce{H}]&={[\ce{A^2-}]}-{[\ce{H2A}]} \\
\mathrm pH &=\frac{\mathrm pK\_\mathrm{a1} + \mathrm pK\_\mathrm{a2} +\log \frac {[\ce{H2A}] + [\ce{H+}] } {[\ce{H2A}]}} {2}\\
\mathrm pH &=\frac{\mathrm pK\_\mathrm{a1} + \mathrm pK\_\mathrm{a2} +\log \left( 1 + \frac { [\ce{H+}] } {[\ce{H2A}]}\right)} {2}\\
K\_\mathrm{1a}&=\ce{[H+][HA-] /[H2A]} \\
\ce{[H2A]}&=\ce{[H+][HA-]} /K\_\mathrm{1a}\\
\mathrm pH &=\frac{\mathrm pK\_\mathrm{a1} + \mathrm pK\_\mathrm{a2} +\log \left( 1 + \frac { K\_\mathrm{1a}} {[\ce{HA-}]}\right)} {2}\\
\end{align}$$


Therefore, if $\mathrm{p}K\_\mathrm{a1}$ is high enough, i.e. if $\ce{H2A}$ is rather a weak acid,  

the simplified formula $\mathrm{p}H=(\mathrm{p}K\_\mathrm{a1} + \mathrm{p}K\_\mathrm{a2})/2$ is right enough. 


If $\mathrm{p}K\_\mathrm{a1}=\ce{[HA-]}$, the $\mathrm{p}H$ correction is cca $+0.15 = \log (2)/2$ 


And then we can use this equation to iterative corrections, with $\ce{[MHA]}$ being concentration of the used acidic salt. The fraction is the classical formula to calculate ion fraction.


$$\ce{[HA-]}=\ce{[MHA]} \cdot \small
\frac {K\_\mathrm{a1}.\ce{[H+]}}
{[\ce{H}]^2+K\_\mathrm{a1} \cdot \ce{[H+]} + K\_\mathrm{a1} \cdot K\_\mathrm{a2}} $$


$$\frac {K\_\mathrm{a1}}{\ce{[HA-]}}=
\frac {[\ce{H}]^2+K\_\mathrm{a1} \cdot \ce{[H+]} + K\_\mathrm{a1} \cdot K\_\mathrm{a2}}
{\ce{[MHA]} \cdot \small
\ce{[H+]}}$$


For particular calculation, the first shoot is taking $\mathrm pH$ from the simplified formula:


$$\begin{align}
\mathrm pH &=\frac{\mathrm pK\_\mathrm{a1} + \mathrm pK\_\mathrm{a2}}{2}=(1+4)/2=2.5 \\
\end{align}$$


I have realized, the direction of iterations must be reversed, as the formulas lead to divergence. ( It is known mathemathical problem to choose the right iteration with the small derivative $\mathrm{d}f(x)/\mathrm{d}x$.


The simplified equation at the bottom of the post expects not too acidic or alkaline $\mathrm pH$. It supposes the equilibrium reactions affecting
 $[\ce{H}]$ do not significantly affect ratio of concentrations $$\frac {[\ce{A^2-}]}{[\ce{H2A}]}$$


As the $\mathrm pH$ is near the both $\mathrm pK\_\mathrm a$, it is quite a fair assumption for most cases.


$$\begin{align}
\ce{2 HA- &<=> H2A + A^2-} \\
\frac {[\ce{H2A}]} {[\ce{HA-}]} &=\frac {[\ce{A^2-}]} {[\ce{HA-}]}\\
 [\ce{H2A}] &= [\ce{A^2-}] \\
\mathrm pH &=\frac{\mathrm pK\_\mathrm{a1} + \mathrm pK\_\mathrm{a2}}{2} \\
\end{align}$$


 7

Brilliant question, and brilliant comment by MaxW, that deserves a non-textbook answer.



> 
> First of all, it implies that the pH is independent of the concentration of HA− present.
> 
> 
> 


Yes, like a buffer.



> 
> Also, if 1/pK1>pK2, then HA− should have the tendency to accept a H+ more than it has of losing one, then why is the pH exactly partway between pK1 and pK2?
> 
> 
> 


The tendency to accept an H+ (as expressed in the ratio of conjugate acid and conjugate base) depends both on the pH and the pKa. If the pH were neutral, we could not have both reactions be at equilibrium. We need a pH that is midway between the pKa values.


Let's treat this as a buffer:


$$\ce{H2A <=> 2 H+ + A^2-}$$


with $K\_{a12} = K\_{a1} \* K\_{a2}.$


Let's check out what happens when $\ce{[H2A] = [A^2-]}$. Inserting the concentrations and taking the negative logarithm, we get:


$$\ce{[H2A]} \times K\_{a12} = \ce{[H+]^2 \times [A^2-]}$$


$$ \mathrm{pK\_{a12} = pK\_{a1} + pK\_{a2} = 2 pH}$$


$$ \mathrm{pH = \frac{pK\_{a1} + pK\_{a2}}{2}}$$



> 
> So why is $\ce{[H2A] = [A^2-]}$?
> 
> 
> 


Well, it works. If you calculate $\ce{[H2A]}$ from pH and $\ce{[HA-]}$, and then calculate $\ce{[A^2-]}$ from pH and $\ce{[HA-]}$, you get the same answer.


If we estimate the pH as:
$$ \ce{[H+]} \overset{!}{=} \sqrt{\mathrm K\_{a1} \times K\_{a2}} $$


For $\ce{[H2A]}$, we get:
$$\ce{[H2A]} \times K\_{a1} = \ce{[H+] \times [AH^-]}$$


$$\ce{[H2A]} = \frac{\ce{[H+]} \times \ce{[AH^-]}}{\mathrm K\_{a1}} = \ce{[AH^-]} \times \sqrt{ \frac{\mathrm K\_{a2}}{\mathrm K\_{a1}}}$$


And for $\ce{[A^2-]}$, we get:
$$\ce{[AH^-]} \times \mathrm{K\_{a2}} = \ce{[H+] \times \ce{[A^2-]}}$$


$$\ce{[A^2-]} = \frac{\mathrm{K\_{a2}} \times \ce{[AH^-]}}{\ce{[H+]}} = \ce{[AH^-]} \times \sqrt{ \frac{\mathrm K\_{a2}}{\mathrm K\_{a1}}}$$


This is a bit of a circular argument, but it is fine because there is always just one equilibrium state for a system. So if you happen to find it and can show everything is at equilibrium, it must be correct.


**How is this different from a buffer?**


The buffer capacity (compared to a conventional 1:1 buffer with a pKa equal to the desired pH) is miserable because the main species is $\ce{HA-}$. You would be able to change the concentration ratio of diprotic acid and the dianion by adding minimal amount of strong acids or base, with a large effect on the pH.


 3

I first found the derivation here
<https://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt>
but the hydrolysis was missing, i tried to derive with hydrolysis and got similar result just with $K\_w$ added.


![enter image description here](https://i.stack.imgur.com/c8QXn.jpg)


$k\_{a1}$, $k\_{a2}$ are first and second dissociation constants of the weak acid $\ce{H2A}$, $k\_w$ is dissociation constant of water  

I have followed equation $\ce{3->1->2}$ (second has the final concentrations boxed)  



![enter image description here](https://i.stack.imgur.com/qkj6K.jpg)


assuming $x, y << C$, $C=[\ce{HA-}]^\circ$ (initial concentration of $\ce{HA-}$)


![enter image description here](https://i.stack.imgur.com/GPA9J.jpg)


the two are almost similar and $k\_w$ can be neglected in many cases (example of $\ce{H3PO3}$ is at the site above, it can be proved by putting $k\_a$ values of $\ce{H3PO3}$, for easy calculation, take  

$pk\_{a1}=2$ and $pk\_{a2}=7$  

and [$\ce{HA-}$] (0.001M will be good for a significant difference in pH) in either of the equations, in second $k\_w$ has to be neglected),


the pH of 0.001 M $\ce{H2PO2-}$ ion must come around 5.2 (assuming third dissociation step is negligible) .  

At $[\ce{HA-}] < 0.01M$ , difference in pH calculated by the above and calculated by  

$pH=\frac{pK\_{a1}+pK\_{a2}}{2}$, increases significantly as $k\_{a1}$ cannot be neglected.


$pH=\frac{pK\_{a1}+pK\_{a2}}{2}$ this can be derived by assuming $k\_{a1} <<[\ce{HA-}]$ in the above equation. Everything is mentioned at the site


 2