1still confused . Please explain clearly . Sorry for the trouble.
18 Answers
1357No, you can assume complete dissociation of HCl at such low concentration.
But you should consider common ion effect to find exact value of [Cl-]
Let the [HCl]=x which gives [H+]=[Cl-]=x(dissociates completely)
[H+]water=10-7
H+ + OH- [eq] H2O
10-7+x 10-7
10-7+x-α 10-7-α
[H+]eq=10-6.5=3.16×10-7=10-7+x-α
(10-7+x-α)(10-7-α)=10-14
3.16×10-7*(10-7-α)=10-14
find α and then find x
1357yeah you are right [Cl-]=[H+]acid
so to calculate [Cl-] you have to find the [H+]acid
but in the question pH is given i.e. [H+]total is given
If we assume [H+]acid=x and [H+]water=y
then [H+]total=x+y=3.16×10-7 (given pH)
also (x+y).y=Kw
so y=10-7/3.16
[Cl-]=[H+]acid=x=3.16×10-7-10-7/3.16
=(3.16-1/3.16)×10-7=2.84×10-7
I think its clear now
1Akshat has asked for [Cl-] , so common ion affects only the dissociation of water , not acid , am I right . So [Cl-] = [H+]acid as if common ion effect was not acting. anyway HCl is a strong acid.
So , if the [H+] is asked then it is equal to the conc due to acid and the net [H+] contribution by water i.e. [H+] in absence of common ion efect minus the correction.
Is this OK?
1357Actually [H+] from water is comparable to [H+] from HCl. So you can't neglect the common ion effect.
As there is little increase in [H+] the equilibrium will shift forward in the above reaction
[H+]=10-7+x-α
For x much greater than 10-7 we can neglect α w.r.t. x
for x much less than 10-7 we can neglect both x and α w.r.t. 10-7
but for comparable conditions we have to consider α
1Manish,,,,
we want ionization constant of HCl to find out the exact value of [Cl-] naa???
plz rep.
1so we want ionization constant of HCl to find out the exact value of [Cl-] naa???
1357I think pH is given in the question which is -log[H+]total
i.e. 3.16×10-7
Let the [HCl]=x which gives [H+]=[Cl-]=x(dissociates completely)
[H+]water=10-7
total [H+]=10-7+x
so definitely x=[Cl-]<[H+]total
If I say pH of 10-8M HCl aqueous solution is close to 6.98
then here [H+]total=10-6.98
but [Cl-]=10-8M<10-6.98
I think its clear now.
1But HCl is strong acid , so it dissociates completely know? Only at high conc it's dissociation is affected . At such low conc , it dissociates completely and also common ion effect can be neglected ,so
[H+]water should be considered. Please correct if wrong.
1357pH is given 6.5, so the [H+]=10-6.5M=3.16×10-7M
But the concentration of HCl is very low, so [H+] from water should be considered.
that means [H+]HCl+[H+]water=3.16×10-7M
which implies [H+]HCl<3.16×10-7M
and so [Cl-]=[H+]HCl<3.16×10-7M
1in aqueos soln . the conc of [H+] = [Cl-] . as it's a monobasic acid.
[H+] = [Cl-] = 10-6.5 = 3.16 x 10 -7 . as the conc . is very low , we must take into account the [H+] comtribution by water [H+] = 10 -7
so . [ H+ ] total = 4.16 x 10 -7