What is the [HF] in 0.02 M HF solution ?

1.	What is the [HF] in 0.02 M HF solution ?
Answer» Solution :`(K_a=3.2xx10^(-4))` HF is weak ACID, its equilibrium established in aqueous solution is as under. `{:(,HF_((aq))+H_2O_((l)) hArr , H_3O_((aq))^(+)+, F_((aq))^(-)),("Initiall",0.02M,0,0),("At equilibrium",(0.02- CALPHA),calpha,C alpha):}` If `alpha`=degree of dissociation there so , decrease in concentration of HF = `c alpha` In solution [HF] =(0.02-0.02 `alpha`)M =0.02 (1-`alpha`) but value of `alpha` is much less , so neglible . `(1-alpha)=1` Thus, [HF]=0.02(1)=0.02 M So, most of weak acid remain as a undissociate molecule in solution and therefore its concentration in solution can be taken of INITIAL concentration C.

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