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The ionization constant of HF is `3.2xx10^(-4)`. Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all the species present `(H_(3)O^(+), F^(-) and HF)` in the solution and its pH .

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`{:((i),HF,+,H_(2)O ,hArr,H_(3)O^(+),+,F^(-)),("Initial conc.",0.02 M ,,,,,,),("Eqm. conc.",0.02-0.02 alpha=0.02(1-alpha),,,,0.02 alpha,,0.02 alpha):}`
`K_(a)=([H_(3)O^(+)][F^(-)])/([HF]):. 3.2 xx 10^(-4)=((0.02 alpha)^(2))/(0.02(1-alpha))=(0.02 alpha^(2))/(1-alpha)~=0.02 alpha^(2)` (Neglecting `alpha` in comparison to 1)
`:. alpha^(2)=(3.2xx10^(-4))/(0.02)=1.6xx10^(-2) or alpha = 0.12`
Note : If `alpha` is not neglected in comparison to 1, solve as follows (solution of quadratic equation) :
`0.02 alpha^(2) = 3.2xx10^(-4) - 3.2 xx 10^(-4) alpha or 2xx10^(-12) alpha^(2) + 3.2 xx 10^(-4) alpha - 3.2xx10^(-4) = 0`
or `alpha^(2) + 1.6xx10^(-2) alpha -1.6xx10^(-2) = 0`
`:. =+ 0.12 and -0.12`.
Neglecting -ve value, `alpha = 0.12`
(ii) Equilibrium concentrations : `[HF]=0.02(1-0.12)=1.76xx10^(-3)M`
`[H_(3)O^(+)]=[F^(-)]=0.02xx0.12=2.4xx10^(-3)M`
(iii) `pH = -log [H_(3)O^(+)]=-log(2.4xx10^(-3))=2.62`

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