1.

Calculate the `pH` of `0.1M K_(3)PO_(4)`soln. The third dissociation constant of orthophoshoric acid is `1.3 xx10^(-12)`. Assume that the hydrolysis proceeds only in the first step.

Answer» First method:
Use direct formula for the `pH` of salt of `S_(B)//W_(A)`
`pH =(1)/(2)(pK_(w) +pK_(a) + logC)`
`= (1)/(2) (14 +11.86 -1) =12.43`
Second method:
`K_(3)PO_(4) +H_(2)O hArr K_(2)HPO_(4) +KOH`
or `PO_(4)^(3-) +H_(2)O hArr HPO_(4)^(2-) + overset(Θ)OH`
Since hydrlysis proceeds only in I step.
`:. [overset(Θ)OH] = C.h =C sqrt(((K_(w))/(K_(a).C)))=sqrt(((K_(w).C)/(K_(a))))`
`K_(a)` is III dissociation constant of acid `H_(3)PO_(4)`
`H_(3)PO hArr H^(o+) +H_(2)PO_(4)^(-1)...K_(1)`
`H_(2)PO_(4)^(Θ) hArr H^(o+) + HPO_(4)^(2-) ...K_(2)`
`HPO_(4)^(2-)hArr H^(o+) +PO_(4)^(3-) K_(3) = K_(a) = 1.3 xx 10^(-12)`
`[overset(Θ)OH] = ((10^(-14)xx0.1)/(1.3xx10^(-12)))^(1//2) = 8.7 xx10^(-2)M`
`:. pOH = 1.5634, pH = 12.4366`


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