1try to use the constrain eq for charges
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