Calculate the pH of a 5.0 M `H_(3)PO_(4)` solution and the equilibrium concentrations of the species `H_(3)PO_(4), H_(2)PO_(4)^(2-) and PO_(4)^(3-). (K_(a_(1))=7.5xx10^(-5)xx10^(-3), K_(a_(2))=6.2xx10^(-8), K_(a_(3))=4.2xx10^(-13))`

Explanation

`{:(,H_(3)PO_(4)+H_(2)O,hArr,H_(3)O^(+),+,H_(2)PO_(4)^(-)),("Initial conc.",5.0 M ,,,,),("Conc. after disso.",5-x,,x,,x),(,,,,,):}`
`K_(a_(1))=([H_(3)O^(+)][H_(2)PO_(4)^(-)])/([H_(3)PO_(4)])=(x.x)/(5-x)=(x^(2))/(5)=7.5xx10^(-3)`
or `x^(2)=37.5xx10^(-3)=3.75xx10^(-2)`
or `x=1.94 xx 10^(-1) = 0.194`
`[H_(3)O^(+)]=[H_(2)PO_(4)^(-)]=0.194M`
`[H_(3)PO_(4)]=5-x=5-0.194=4.806M`
As `K_(a_(2)) and K_(a_(3))` are `lt lt K_(a_(1))`, hence `[H_(3)O^(+)]` is mainly from this step. Hence,
`pH = - log [H_(3)O^(+)]=- log 0.194 = 0.71`
Step 2 of dissociation :
`{:(,H_(2)PO_(4)^(-)+,H_(2)O,hArr,H_(3)O^(+)+,HPO_(4)^(2-)),("Initial",0.194M,,,,),("After dissociation",0.194-y,,,0.194+y,y),(,,,,~=0.194,):}`
`K_(a_(2))=(0.194xxy)/(0.194-y)~=(0.194y)/(0.194)=y=6.2xx10^(-8)`
Hence, `[HPO_(4)^(2-)]=6.2xx10^(-8)`
Step 3 of dissociation :
`{:(,HPO_(4)^(2-)+,H_(2)O,hArr,H_(3)O^(+)+,PO_(4)^(3-)),("Initial",6.2xx10^(-8)M,,,0.194+z,z),("After disso.",6.2xx10^(-8)-z,,,~=0.194,),(,,,,,):}`
`K_(a_(3))=(0.194xxz)/(6.2xx10^(-8)-z)~=(0.194z)/(6.2xx10^(-8))=4.2xx10^(-13)`
or `z=(4.2xx10^(-13)xx6.2xx10^(-8))/(0.194)=134xx10^(-21)=1.34xx10^(-19)`
`:. [PO_(4)^(3-)]=1.34xx10^(-19)` (negligible)
Note that in the 2nd step of dissociation, concentration of anion is equal to `K_(a_(2))` value.