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

Explanation

`{:("Step I",H_(3)PO_(4),hArr,H^(oplus),+,H_(2)PO_(4)^(Theta) ,K_(1)=7.5 xx 10^(-3)),("Step II",H_(2)PO_(4)^(Theta),hArr,H^(oplus),+,HPO_(4)^(2-) , K_(2)=6.2 xx 10^(-4)),("Step III",HPO_(4)^(2-),hArr,H^(oplus),+,PO_(4)^(3-) ,K_(3)=3.6 xx10^(-13)),("For Step I",H_(3)PO_(4),hArr,H^(oplus),+, H_(2)PO_(4)^(Theta)),(,0.1,,0,,0),(,0.1-C,,C,,C):}`
`K_(1) = ([H^(o+)][H_(2)PO_(4)^(Θ)])/([H_(3)PO_(4)]) = (C.C)/((0.1-C))`
`7.5 xx 10^(-3) = (C^(2))/((0.1 - C))`
`:. C = 0.024` and `[H^(o+)] = 0.024M, [H_(2)PO_(4)^(Θ)] = 0.024M, [H_(3)PO_(4)] = 0.1 - 0.024 = 0.76M`
The value fo `K_(1)`is much large than `K_(2)` and `K_(3)`.
Also dissociation of steps `II` and `III` occurs in presence of `H^(o+)` furnished in step `I` and thus, dissociation of steps `II` and `III` is further suppressed due to common effect.
`{:("For Step II :",H_(2)PO_(4)^(Theta),hArr,H^(oplus)+,H_(2)PO_(4)^(2-)),(,0.024,,0.024,0),(,(0.024-y),,(0.024+y),y):}`
The dissociation of `H_(2)PO_(4)^(Θ)` occurs in presence of `[H^(o+)]` furnished in step I.
Thus, `K_(2) =([H^(o+)][HPO_(4)^(2-)])/([H_(3)PO_(4)])`
or `6.2 xx 10^(-8) = ((0.024 +y)y)/((0.024-y))` because `y` is small.
Assuming `0.024 - y ~~ 0.024`and neglecting `y^(2)`
`:. 6.2 xx 10^(-8) = (0.024y)/(0.024)`
`:. y = 6.2 xx 10^(-8)`
or `[HPO_(4)^(2-)] = K_(2) = 6.2 xx 10^(-8)` [Insignificant]
`{:("For Step III :",HPO_(4)^(2-),hArr,H^(oplus),+,PO_(4)^(3-)),(,(6.2 xx10^(-8)-x),,(0.024+x),,x):}`
`:.K_(3) = ([H^(o+)][PO_(4)^(3-)])/([HPO_(4)^(2-)]) = ((0.024 +x)x)/([(6.2 xx 10^(-8)-x)])`
Again neglecting `x^(2)` and assuming `(6.2 xx 10^(-8)) - x ~~ 6.2 xx 10^(-8)`
`:. 3.6 xx 10^(-13) = (0.024x)/(6.2 xx 10^(-8))`
`:. x = (3.6 xx 106(-13) xx 6.2 xx 10^(-18))/(0.024) = 9.3 xx 10^(-19) `[Insignificant]