Question

0.0001 N HCl विलयन की pH होगी​

Answer 1

pH = -log[$H^{+}$]

HCl is a strong electrolyte

It would completely dissociate

So [HCl] = [$H^{+}$]

Thus [$H^{+}$] = 0.0001

[$H^{+}$] = $10^{-4}$

-log[$H^{+}$] = -log $10^{-4}$ = pH = 4

Thus pH = 4

since Hydrogen ion coming from HCl is in significant quantity we can neglect Hydrogen ion coming from Water.

Might Help Thanks