How do I find the pH of a diprotic acid?

In fact you want #pK_(a1)# and #pK_(a2)# for sulfuric acid and bisulfate ion... Sulfuric acid is an exceptionally strong acid, and may be hard to measure in water...i.e. the first protonolysis reaction may lie strongly to the right....

#H_2SO_4(aq) + H_2O(l) rightleftharpoonsH_3O^+ + HSO_4^(-)#

...the second equivalent of protium is still accessible from its conjugate base, the #"bisulfate ion"#...

#HSO_4^(-) + H_2O(l) rightleftharpoonsSO_4^(2-) + H_3O^+#

This site quotes #pK_(a2)=1.99#...and #pK_(a1)=-3#

Sulfuric acid is diprotic which means it has two hydrogen ions which may be given up:

#sf(H_2SO_4rarrHSO_4^(-)+H^+" "color(red)((1))#

#sf(K_(a1)# is very large.

#sf(HSO_4^(2-)rightleftharpoonsSO_4^(2-)+H^+" "color(red)((2))#

#sf(K_(a2)=([SO_4^(2-)][H^+])/([HSO_4^-])=0.01)#

The 1st ionisation can be considered to be just about 100% complete.

For the 2nd ionisation Le Chatelier's Principle tells us that this process will be suppressed by the presence of the #sf(H^+)# ions from #sf(color(red)((1))#.

For this reason you tend to find that #sf(K_a)# values for subsequent ionisations for polyprotic acids are often smaller.

The #sf(H^+)# is also reluctant to leave a negatively charged species which is the case here.

In your comment you ask how to find the pH of a 0.5 M solution of #sf(H_2SO_4)# so here's how you could do this:

Since #sf(color(red)((1))# has gone to completion we can assume that the initial concentrations of #sf(HSO_4^-)# and #sf(H^+)# must also be 0.5 M . We can put this onto an ICE table based on concentration in M :

#sf(color(white)(xxxxx)HSO_4^(2-)rightleftharpoonsSO_4^(2-)+H^+)#

#sf(Icolor(white)(xxxxxx)0.5color(white)(xxxxx)0color(white)(xxxx)0.5)#

#sf(Ccolor(white)(xxxx)-xcolor(white)(xxxx)+xcolor(white)(xxx)+x)#

#sf(Ecolor(white)(xxxx)(0.5-x)color(white)(xxx)xcolor(white)(xxx)(0.5+x))#

#:.##sf(K_(a2)=(x(0.5+x))/((0.5-x))=0.01)#

To solve for x we could multiply out the brackets leaving us with a quadratic equation to solve for x . To make things easier I will assume that x is sufficiently small to make the assumption that:

#sf((0.5+x)rArr0.5)#

and

#sf((0.5-x)rArr0.5)# (I will test this later).

#:.##sf((xcancel(0.5))/(cancel(0.5))=0.01)#

#sf(x=0.01)#

From the ICE table we can say that :

#sf([H^+]=(0.5+x)=(0.5+0.01)=0.51color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log[0.51]=0.29)#

Normally a value of 0.01 for #sf(K_(a2))# would be too large to make the "small x" approximation. However I will plug it back into the expression for #sf(K_(a2)rArr)#

#sf(0.01xx((0.5+0.01))/((0.5-0.01))=0.0104)#

which is very close to 0.01 - the value of #sf(K_(a2))#