How do you calculate pH diprotic acid?

Feb 15, 2014

I would not typically teach this to my high school students, so I looked around and found a great explanation on you tube.
Since, in a polyprotic acid the first hydrogen will dissociate faster than the others, If the Ka values differ by a factor of 10 to the third power or more, it is possible to approximately calculate the pH by using only the Ka of the first hydrogen ion. For example:

Pretend that ${H}_{2} X$ is a diprotic acid. Look up on a table the Ka1
for the acid.

If you know the concentration of the of the acid, say it is 0.0027M and the $K {a}_{1}$ is $5.0 x {10}^{- 7}$. Then you can set up your equation as follows;
${H}_{2} X$ --> ${H}^{+ 1}$ + $H {X}^{- 1}$ with $K {a}_{1}$ = $5.0 x {10}^{- 7}$ Using the formula: Ka=(products) / (reactants):

$5.0 x {10}^{- 7}$ = (${x}^{2}$)/(0.0027) Then solve for x and you have your hydrogen ion concentration. Since pH = -log of hydrogen ion concentration, you can now calculate the pH.