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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 species present (H_3O^(+), F^- and HF ) in the solution and its pH. |
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Answer» Solution :In aqueous solution of HF, two proton transfer reactions occurs like, (i)Ionization of HF (ii) Self ionization of water. (i)`HF_((aq))+H_2O_((ll)) hArr H_3O_((aq))^(+) + F_((aq))^(-)` (ii)`H_2O_((l)) + H_2O_((l)) hArr H_3O_((aq))^(+)+ OH_((aq))^(-)` `K_a (i) =3.2xx10^(-4)` and `K_w`(ii) = `1.0xx10^(-14)` At `K_a (i) gt gt K_w (ii)`, So, main EQUILIBRIUM in solution is ionization of HF. Suppose , dissociation amount of HF= `alpha` Initial [HF]=0.02 M `therefore` Dissociated [HF]=`(0.02alpha)M` ` therefore` At equilibrium [HF]=(Initial - Dissociated ) =(0.02 - 0.02 `alpha`) `=0.02 (1-alpha)M` `H_3O^+` and `F^-` produce equal to the amount of the HF dissociate . `therefore` At equilibrium `[H_3O^+]=[F^-] 0.02 alpha M` These are SHOWN in the following table. `{:("Equilibrium :",HF_((aq))+,H_2O_((l)) hArr, H_3O_((aq))^(+) +, F_((aq))^(-)),("Initial M:", 0.02, -,Zero,Zero),("Change:", -0.02alpha,-,+0.02alpha,+0.02 alpha),("Equili (M)", 0.02 - 0.02 alpha, , 0.02 alpha,0.02 alpha):}` Equilibrium constant for ionization of HF. `K_a=([H_3O^+][F^-])/"[HF]"` `therefore 3.2 xx10^(-4) = ((0.02 alpha)(0.02alpha))/(0.02 (1-alpha))=(0.02 alpha^2)/(1-alpha)` `therefore 3.2xx10^(-4) (1-alpha)=0.02 alpha^2` `therefore 0.02 alpha^2 + 3.2 xx10^(-4) alpha -3.2 xx10^(-4) = 0.0` `therefore alpha^2 + 0.016 alpha - 0.016 =0.0` For this quadratic equation a=1,b=0.016 and c=-0.16 `therefore alpha =(-b+sqrt(b^2-4ac))/(2a)` `=(-0.016 pm sqrt((0.016)^2 - 4(1)(-0.016)))/(2(1))` `=(-0.016 pm sqrt(0.000256+0.064))/2` `=(-0.016 pm 0.2535)/2` `=0.2375/2` `-0.2695/2` =0.1187 OR -0.1347 `approx` Dissociation degree or negative value is not possible . At equilibrium `[H_3O^+] = [F^-]=0.02alpha` `=0.02xx0.1187` = 0.002374 `=2.374xx10^(-3)` M `approx 2.4 XX 10^(-3)` M At equilibrium [HF]=`(0.02-0.02 alpha)` =0.02-0.02(0.12)=(0.02-0.0024) =0.0176 `=1.76xx10^(-2)` M pH=log `[H_3^+O]`=log `(2.4xx10^(-3))` =-(0.3802-3)=-(-2.6198) =+2.6198 `approx` 2.62 OR `K_a=(H_3O^+)^2/C_o=3.2xx10^(-4)` `[H_3O^+]=sqrt(K_aC_o)=sqrt(3.2xx10^(-4)xx0.02)` `=sqrt(6.4xx10^(-6))= 2.53xx10^(-3)` M So, `[H_3O^+]=[F^-]=2.53xx10^(-3)` M = 0.00253 M [HF]=(0.02-0.00253)=0.01747 `approx` 0.017 M pH=-log `(2.53xx10^(-3))` = 2.5969 `approx` 2.6 |
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