Question

Sir/madam,

I have two questions which are almost similar but answers are given in two different ways..

questions are:Q1)Assuming complete dissociation calculate pH for 3×10^-3M HCl?

Q2)Calculate pH of a 1×10^-8M of HCl?

I know the answers of both questions but they are done in different ways..So is there any difference?

Answer 1

1.) Formulas to be used:
[OH-] = Molarity × Acidity
[H+] = Molarity × Basicity
pH = -log[H+]
pOH = -log[OH-]
pH = 14 - pOH

(a)
[H+] = 3 × 10^-3
pH = -log[3 × 10^-3]
pH = 3 - log3
pH = 2.52



2.) If we use the relation, pH = – log [H3O+], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when H+ concentrations from acid and water are comparable, the concentration of H+ from water cannot be neglected.

Therefore,
[H+] total = [H+] acid + [H+] water
Since HCl is a strong acid and is completely ionized
[H+] HCl = 1.0 x 10-8
The concentration of H+ from ionization is equal to the [OH–] from water,
[H+] H2O = [OH–] H2O
= x (say)
[H+] total = 1.0 x 10-8 + x
But
[H+] [OH–] = 1.0 x 10-14
(1.0 x 10-8 + x) (x) = 1.0 x 10-14
X2 + 10-8 x – 10-14 = 0
Solving for x, we get x = 9.5 x 10-8
Therefore,
[H+] = 1.0 x 10-8 + 9.5 x 10-8
= 10.5 x 10-8
= 1.05 x 10-7
pH = – log [H+] = – log (1.05 x 10-7) = 6.98
& So 2 of the above questions are answered in 2 different ways