A little confused on how to calculate the PH off of this.
Strong acid completely dissociating
Hi!
n=C * V
C = n/V = 10^-4 / 10^-1 L = 10^-3M
HCl + H2O = H3O+ + Cl-
1:1 ratio HCL and H3O+ so also 10^-3M of H3O+
pH = - log [H+] = - (-3)= 3
Edit: typo
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