Question

Calculate the pH of a Solution which Contain 5 * 10^-8 moles of Hcl in 1 litre of water at 25 C

Answer 1

@Kryten @Aperogalics

Answer 2

c= n/V = 5*10^-8 mol/1dm3 = 5*10^-8 M
pH = - log [H+]
pH = - log(5*10^-8)

Answer 3

I'm afraid Kryten answered too quickly. The result would be greater than 7, which is not possible for an acidic solution.

This is because the H3O+ ions brought by water itself cannot be neglected.

Actually, you need to write down water self-ionization reaction and work out the amounts in solution so that they match water's ionic product.

Answer 4

Answer should be pH = 6.89

Answer 5

Yup u r correct @Vincent-Lyon.Fr but Hw u Did it ?

Answer 6

Can you do this ?
"write down water self-ionization reaction and work out the amounts in solution"

Use notations:
\(K_w\) water self-ionization constant
\(c_o\) concentration of the strong acid
h = \([H_3O^+]\)
ω = \([OH^-]\)

Answer 7

kw = [H+] [0H-]

Answer 8

OK!
Now, write down amounts in solution (1 litre) using the reaction's stoichiometry.

Answer 9

HCl + H20 ----> H30 + + Cl-

Answer 10

Ok,
This reaction is not an equilibrium and leads to an initial amount of \(c_o\) mole of \(H_3O^+\) in water.

Now write down the relevant equation, which is that of the reaction undergone by water.