365 gr HCL (Mr=36,5) dilarutkan dalam air hingga 500 ml larutan,berapa Ph larutan tsb

HCl —-> H+ + Cl-

nHCl = gr/Mr

        = 365/36.5

        = 10 mol

[H+] = n/V(l)

       = 10/0.5

       = 2 x 10^1

pH= -log[H+]

    = -log[2 x 10^1]

   = -1-log2