1.

Find the degree of dissociation of HF in 1 M aqueous solution. The value of K for the ionic equilibrium HF=H^(+)+F^(-) is 7.2xx10^(-4).

Answer»

Solution :HF dissociates in water to form `H^(+)` and `F^(-)` ions. On reaching the equilibrium we have
`HF HARR H^(+)+F^(-)`
Thus one mole of HF taken initially dissociates to yield 1 mole of `H^(+)` and 1 mole of `F^(-)`.
If x be the degree of DISSOCIATION, the concentration TERMS at equilibrium are :
`[HF]=(1-x)"mol/L"`
`[F^(-)]=xmol//L`
`[H^(+)]=x mol//L`
Substituting these values in the equilibrium expression, we have
`K=7.2xx10^(-4)=([H^(+)][F^(-)])/([HF])=((x)/(x))/((1-x))`
If x is very small COMPARED to 1, we can write :
`7.2xx10^(-4)=(x^(2))/(1.00)`
`thereforex=(7.2xx10^(-4))^(1//2)`
Thus the degree of dissociation of HF in 1 M solution is `2.7xx10^(-2)`


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