3.) For a 0.12 M solution of HCl (aq), calculate the following: a. [H3O+] b. pH c. pOH d. [OH-]
Given Conc. of HCl = 0.12 M
So;
a) [H3O+] = 0.12
b) pH = - log[H+] = - log [0.12] = 0.92
c) pH + pOH = 14; pOH = 14 - pH = 14 - 0.92 = 13.08
d) [OH-] = 10-pOH = 10-13.08 = 8.3 x 10-14
3.) For a 0.12 M solution of HCl (aq), calculate the following: a. [H3O+] b. pH...
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