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jml mol H+ pd larutan A = 0,1 . 100ml = 10 mmol
pH = 2 maka [H+] = 10^-2 = 0,01 M
jml mol H+ pd lar. B = 0,01 . 100ml = 1 mmol
[H+] = total n H+/ total V = (10+1)mmol/(100+100)ml = 11/200 M = 0,055 M
pH = -log 0,055 = 1,23
maka [H+] = 10^-1 = 0,1 M
jumlah mol H+ pada larutan A = 0,1 x 100ml = 10mmol
pH2 = 2 maka [H+] = 10^-2 = 0,01M
jumlah mol H+ pada larutan B = 0,01 x 100ml = 1mmol
[H+] = total n H+/total V = (10+1) mmol / (100+100) ml = 11/200 M = 0,055 M
pH = -log 0,055 = 1,23
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