InterviewSolution
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InterviewSolution
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The ionisation constant of `HF` is `3.2 xx 10^(-4)`. a. Calculate the dergee of dissociation of aall species present `M` solution. b. Calculate the concentration of all species present `(H_(3)O^(o+),F^(Θ)` and `HF`) in the solution. c. Calculate method: |
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Answer» a. The following proton transfter reactions are possible, `HF +H_(2)O hArr H_(3)O^(oplus)+F^(Theta)(K_(a)=3.2 xx10^(-4))` ....(i) `H_(2)O+H_(2)O hArr H_(3)O^(oplus)+overset(Theta)(O)H (Kw=10^(14))`....(ii) Since `K_(a)gt gt K_(w)`, reaction (i) is the main reaction. `{:(,HF+, H_(2)O hArr,H_(3)O^(oplus),F^(Theta)),("Initial conc",rArr0.02,-,0,0),("Change in conc",rArr-0.02alpha,-,+0.02alpha,+0.02alpha),("Eq conc",rArr0.02-0.02alpha,-,,),(,=0.021(1-alpha),,0.02alpha,0.02alpha):}` `:.K_(a) = ((0.02alpha)^(2))/(0.02(1-alpha)) = (0.02alpha^(2))/(1-alpha) = 3.2 xx 10^(-4) ..(i)` Solving the equaiton (i), the following quadratic equaiton is obained, `alpha^(2) +1.6 xx 10^(-2)alpha - 1.6 xx 10^(-2) = 0`. `:. alpha = +0.12` and `-0.12`. (-ve value is not taken) Thus, `alpha = 0.12` b. `[H_(3)O^(o+)] = [F^(Θ)] = C alpha = 0.02 xx 0.12 = 2.4 xx 10^(-3) M` `[HF] = C(1-alpha) = 0.02 (1-0.12) = 17.6 xx 10^(-3)M` c. `pH = - log [H^(o+)] =- log (2.4 xx 10^(-3)) = 2.62` Note: `alpha - 0.12` which is greater than `0.05, (alpha ge 0.05` or `alpha le 5%)` So the term `(1-alpha) != 1`, is taken. Second method: By using formula of `alpha`, and `ph`. a. `alpha = sqrt((K_(a))/(C)) = sqrt((3.2 xx 10^(-4))/(0.02)) = sqrt(16xx10^(-4)xx10)` `= 4 xx 10^(-2) xx sqrt(10)` `(sqrt(10) = 3.162)` `= 4xx3.162 xx 10^(-2)` `= 0.126` `:. alpha= 0.12` b. `[H_(3)O^(o+)] = [F^(Θ)] = Calpha = 0.02 xx 0.12 = 2.4 xx 10^(-3)M` `[HF] = C(1-alpha) = 0.02 (1-0.12) = 1.76 xx 10^(-3)M` c. `pK_(a) =- log (3.2 xx 10^(-4)) =- log (32 xx 10^(-5))` `=- log (25 xx 10^(-5))` `=- 5 log 2+5` `=- 5 xx 0.3 +5 = 3.5` `pK_(a) ~~ 3.5` `pH = (1)/(2) (pK_(a) - logC) = (1)/(2) (3.5-log0.02)` `= (1)/(3) (3.5 - log 2 xx 10^(-2))` `= (1)/(2) (3.5 - 0.3 +2)` `= (1)/(2) (5.2) = 2.6` `:. pH ~~ 2.6`. |
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