Question

Calculate the ph of 1×10-8m Hcl solution

Answer 1

Answer:

Since it is very dilute acidic solution, so H+ concentrations from acid and water are comparable, nd 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 (assume)

[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

X^2 + 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

✨Hope it will be helpful.✨