1.

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

Answer» Strategy: Identify the primary reaction and write the reaction summary in terms of unknwon x. Solve for x, knowing the value of `K_a`.
Solution:
Step 1-2: The following ionization reactions are possible:
`HF(aq.)=H^(+)(aq.)+F^(-)(aq.)`
`K_a=3.2xx10^(-4)`
`H_2O(1)=H^(+)(aq.)+OH^(-)(aq.)`
`K_w=1.0xx10^(-14)`
Step 3: As `K_w lt lt K_a`, the first reaction is the primary reaction.
Step 4: Let x be the equilibrium concentration of `H^(+)` and `F^(-)` ions in `mol L^(-1)`. Then the equilibrium concentration of `HF` must be `(0.02-x)mol L^(-1)`. We can write the reaction summary as follows:
`{:(,HF(aq.)hArrH^(+)(aq.)+F^(-)(aq.)),("Initial (M)"," 0.02 0 0"),("Change (M)"," -x +x +x"),("Equilibrium (M)",bar("(0.02-x) x x ")):}`
Step 5:
`K_a=(C_(H^+)C_(F^-))/(C_(HF))`
`3.2xx10^(-4)=((x)(x))/(0.02-x)=(x^2)/(0.02-x)`
This equation can be written as
`x^2+3.2xx10^(-4)x-6.4xx10^(-6)=0`
which fits the quadratic equation `ax^2+bx+c=0`. Using the quadratic formula
`x=(-b+-sqrt(b^2-4ac))/(2a)`
we get
`x=2.4xx10^(-3)M` or `-2.4xx10^-3M`
The second solution is physically impossible since the concentration of ions produced as a result of ionization cannot be negative.
Thus,
`x=2.4xx10^(-3)M`
Step 6: Thus, at equilibrium,
`C_(H^+)=2.4xx10^(-3)M`
`C_(F^-)=2.4xx10^(-3)M`
`C_(HF)=(0.02-2.4xx10^(-3))M`
`=1.76xx10^(-2)M`
`alpha=(C_(H^+))/(C_(HF))=(2.4xx10^(-3))/(0.02)=0.12`
`pH=-"log"(C_(H^+))/(mol L^(-1))=-log(2.4xx10^(-3))`
`=2.62`


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