Calculate the concentration of all species of significant concentrations presents in `0.1 M H_(3)PO_(4)` solution. lf `K_(1) = 7.5 xx 10^(-3), K_(2) = 6.2 xx 10^(-8), K_(3) = 3.6 xx 10^(-13)`

Explanation

As the value of `K_1` is much larger than `K_2 and K_3` it may be assumed that Ht formed in the second and third steps of dissociation of `H_3PO_4` is insignificant. Further, as `H_2O` dissociates feebly, `H^+` ions produced by the dissociation of `H_2O` may also be neglected. The species of significant concentration are `H_3PO_4, H^+ and H_2PO_4^-.` Now, first dissociation of `H_3PO_4` is
`overset(0.1)(H_(3)PO_(4)) Leftrightarrow H_(2)PO_(4)^(-)+H^+`
`"(0.1-x) x x Concentration at eqb."`
`K_(1)=(x.x)/(0.1-x)=7.5 xx 10^(-3), x=0.024`
Thus, from the first step of dissociations `[H^+]=[H_(2)PO_(4)^(-)]=0.024M`
and `[H_(3)PO_(4)]=0.1-0.024=0.076M`
For the second dissociation, initial concentration of `H_2PO_4^(-)` is 0.024 M
`underset((0.24-y))(H_(2)PO_(4)^(-)) Leftrightarrow underset(y)(HPO_(4)^(-))+underset(y+0.024)(H^+)` Concentration at eqb.
`K_(2)=((y(y+0.024))/(0.024-y))=6.2 xx 10^(-8)`
As `0.024 gt gt y (0.024y)/(0.024)=6.2 xx 10^(-8), y=6.2 xx 10^(-8)`
`[HPO_(4)^(2-)]=K_(2)`
Thus, 0.1 `M H_(3)PO_(4)" contains "[H_(3)PO_(4)]=0.076M`
`[H^+]=y+0.024 approx 0.024M`
`[H_(2)PO_(4)^(-)]=0.024 -y approx 0.024M`